Class 11: Hyperbola – Exercise 27.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: The equation of the directrix of a hyperbola is $\displaystyle x-y + 3-0.$ Its focus is $\displaystyle (- 1, 1)$ and eccentricity $\displaystyle 3$ . Find the equation of the hyperbola.

Answer:

Let $\displaystyle S(-1, 1)$ be the focus and $\displaystyle P(x, y)$ be a point on the Hyperbola.

Draw $\displaystyle PM$ perpendicular from $\displaystyle P$ on the directrics. Then, by definition

$\displaystyle SP = e PM$

$\displaystyle \Rightarrow SP^2 = e^2 PM^2$

$\displaystyle \Rightarrow (x+1)^2 + ( y - 1)^2 = (3)^2 \Bigg[ \frac{x-y+3}{\sqrt{(1)^2+(-1)^2}} \Bigg]^2 \hspace{2.0cm} [\because e = 3]$

$\displaystyle \Rightarrow x^2 + 1 + 2x + y^2 + 1 - 2y = \frac{9}{2} \Big[ x-y+3 \Big]^2$

$\displaystyle \Rightarrow 2[ x^2 + y^2 + 2x - 2y + 2 ] = 9 [ x-y+3]^2$

$\displaystyle \Rightarrow 2x^2 + 2y^2 + 4x - 4y + 4 = 9[ x^2 + y^2 + 9 -2xy-6y+6x ]$

$\displaystyle \Rightarrow 2x^2 + 2y^2 + 4x - 4y + 4 = 9x^2+9y^2+81-18xy-54y+54x$

$\displaystyle \Rightarrow 7x^2 + 7y^2-18xy+50x-50y+77=0$

This is the required equation of the hyperbola.

$\\$

Question 2: Find the equation of the hyperbola whose

(i) focus is $\displaystyle (0, 3),$ directrix is $\displaystyle x + y - 1 = 0$ and eccentricity $\displaystyle = 2$

(ii) focus is $\displaystyle (1, 1),$ directrix is $\displaystyle 3 x + 4y + 8=0$ and eccentricity $\displaystyle =2$

(iii) focus is $\displaystyle (1, 1)$ directrix is $\displaystyle 2x+y =1$ and eccentricity $\displaystyle = \sqrt{3}$

(iv) focus is $\displaystyle (2,-1),$ directrix is $\displaystyle 2 x + 3 y =1$ and eccentricity $\displaystyle =2$

(v) focus is $\displaystyle (a,0),$ directrix is $\displaystyle 2 x -y + a=0$ and eccentricity $\displaystyle = 4/3$

(vi) focus is $\displaystyle (2,2),$ directrix is $\displaystyle x + y = 9$ and eccentricity $\displaystyle = 2$

Answer:

(i) Given focus is $\displaystyle (0, 3),$ directrix is $\displaystyle x + y - 1 = 0$ and eccentricity $\displaystyle = 2$

Let $\displaystyle S(0, 3)$ be the focus and $\displaystyle P(x, y)$ be a point on the Hyperbola.

Draw $\displaystyle PM$ perpendicular from $\displaystyle P$ on the directrics. Then, by definition

$\displaystyle SP = e PM$

$\displaystyle \Rightarrow SP^2 = e^2 PM^2$

$\displaystyle \Rightarrow (x-0)^2 + ( y - 3)^2 = (2)^2 \Bigg[ \frac{x+y-1}{\sqrt{(1)^2+(1)^2}} \Bigg]^2 \hspace{2.0cm} [\because e = 2]$

$\displaystyle \Rightarrow x^2 + y^2 + 9 - 6y = \frac{4}{2} \Big[ x+y-1 \Big]^2$

$\displaystyle \Rightarrow x^2 + y^2 - 6y + 9 = 2 [ x+y-1]^2$

$\displaystyle \Rightarrow x^2 + y^2 - 6y + 9 = 2[ x^2 + y^2 + 1 +2xy-2y-2x ]$

$\displaystyle \Rightarrow x^2 + y^2 - 6y + 9 = 2x^2 + 2y^2 + 2 +4xy-4y-4x$

$\displaystyle \Rightarrow x^2 + y^2+4xy-4x+2y-7=0$

This is the required equation of the hyperbola.

$\\$

(ii) Given focus is $\displaystyle (1, 1),$ directrix is $\displaystyle 3 x + 4y + 8=0$ and eccentricity $\displaystyle =2$

Let $\displaystyle S(1, 1)$ be the focus and $\displaystyle P(x, y)$ be a point on the Hyperbola.

Draw $\displaystyle PM$ perpendicular from $\displaystyle P$ on the directrics. Then, by definition

$\displaystyle SP = e PM$

$\displaystyle \Rightarrow SP^2 = e^2 PM^2$

$\displaystyle \Rightarrow (x-1)^2 + ( y - 1)^2 = (2)^2 \Bigg[ \frac{3x+4y+8}{\sqrt{(3)^2+(4)^2}} \Bigg]^2 \hspace{2.0cm} [\because e = 2]$

$\displaystyle \Rightarrow x^2 + 1 - 2x + y^2 + 1 - 2y = \frac{4}{25} \Big[3x+4y+8 \Big]^2$

$\displaystyle \Rightarrow 25[ x^2 + y^2 - 2x - 2y + 2 ] = 4 [ 3x+4y+8]^2$

$\displaystyle \Rightarrow 25x^2 + 25y^2 -50x - 50y + 50 = 4[ 9x^2 + 16y^2 + 64 +24xy+64y+48x ]$

$\displaystyle \Rightarrow 25x^2 + 25y^2 -50x - 50y + 50 = 36x^2 + 64y^2 + 256 +96xy+256y+192x$

$\displaystyle \Rightarrow 11x^2 + 39y^2+96xy+242x+306y+206=0$

This is the required equation of the hyperbola.

$\\$

(iii) Given focus is $\displaystyle (1, 1)$ directrix is $\displaystyle 2x+y =1$ and eccentricity $\displaystyle = \sqrt{3}$

Let $\displaystyle S(1, 1)$ be the focus and $\displaystyle P(x, y)$ be a point on the Hyperbola.

Draw $\displaystyle PM$ perpendicular from $\displaystyle P$ on the directrics. Then, by definition

$\displaystyle SP = e PM$

$\displaystyle \Rightarrow SP^2 = e^2 PM^2$

$\displaystyle \Rightarrow (x-1)^2 + ( y - 1)^2 = (\sqrt{3})^2 \Bigg[ \frac{2x+y-1}{\sqrt{(2)^2+(1)^2}} \Bigg]^2 \hspace{2.0cm} [\because e = 2]$

$\displaystyle \Rightarrow x^2 + 1 - 2x + y^2 + 1 - 2y = \frac{3}{5} \Big[2x+y-1 \Big]^2$

$\displaystyle \Rightarrow 5[ x^2 + y^2 - 2x - 2y + 2 ] = 3 [ 2x+y-1]^2$

$\displaystyle \Rightarrow 5x^2 + 5y^2 -10x - 10y + 10 = 3[ 4x^2 + y^2 + 1 +4xy-2y-4x ]$

$\displaystyle \Rightarrow 5x^2 + 5y^2 -10x - 10y + 10 = 12x^2 + 3y^2 + 3 +12xy-6y-12x$

$\displaystyle \Rightarrow 7x^2 -2y^2+12xy-2x+4y-7=0$

This is the required equation of the hyperbola.

$\\$

(iv) Given focus is $\displaystyle (2,-1),$ directrix is $\displaystyle 2 x + 3 y =1$ and eccentricity $\displaystyle =2$

Let $\displaystyle S(2, -1)$ be the focus and $\displaystyle P(x, y)$ be a point on the Hyperbola.

Draw $\displaystyle PM$ perpendicular from $\displaystyle P$ on the directrics. Then, by definition

$\displaystyle SP = e PM$

$\displaystyle \Rightarrow SP^2 = e^2 PM^2$

$\displaystyle \Rightarrow (x-2)^2 + ( y +1)^2 = (2)^2 \Bigg[ \frac{2 x + 3 y - 1}{\sqrt{(2)^2+(3)^2}} \Bigg]^2 \hspace{2.0cm} [\because e = 2]$

$\displaystyle \Rightarrow x^2 + 4 -4x + y^2 + 1 + 2y = \frac{4}{13} \Big[ 2 x + 3 y - 1 \Big]^2$

$\displaystyle \Rightarrow 13[ x^2 + y^2 -4x + 2y + 5 ] = 4 [2 x + 3 y - 1]^2$

$\displaystyle \Rightarrow 13x^2 + 13y^2 -52x + 26y + 65 = 4[ 4x^2 + 9y^2 + 1 +12xy-6y-4x ]$

$\displaystyle \Rightarrow 13x^2 + 13y^2 -52x + 26y + 65 = 16x^2+36y^2+4+48xy-24y-16x$

$\displaystyle \Rightarrow 3x^2 + 23y^2+48xy+36x-50y-61=0$

This is the required equation of the hyperbola.

$\\$

(v) Given focus is $\displaystyle (a,0),$ directrix is $\displaystyle 2 x -y + a=0$ and eccentricity $\displaystyle = 4/3$

Let $\displaystyle S(a, 0)$ be the focus and $\displaystyle P(x, y)$ be a point on the Hyperbola.

Draw $\displaystyle PM$ perpendicular from $\displaystyle P$ on the directrics. Then, by definition

$\displaystyle SP = e PM$

$\displaystyle \Rightarrow SP^2 = e^2 PM^2$

$\displaystyle \Rightarrow (x-a)^2 + ( y - 0)^2 = (\frac{4}{3})^2 \Bigg[ \frac{2 x -y + a}{\sqrt{(2)^2+(-1)^2}} \Bigg]^2 \hspace{2.0cm} \Big[ \because e = \frac{4}{3} \Big]$

$\displaystyle \Rightarrow x^2 + a^2 -2ax + y^2 = \frac{16}{45} \Big[2 x -y + a \Big]^2$

$\displaystyle \Rightarrow 45[ x^2 + y^2 - 2ax +a^2 ] = 16 [ 2 x -y + a ]^2$

$\displaystyle \Rightarrow 45x^2 + 45y^2 - 90ax +45a^2 = 16[ 4x^2 + y^2 + a^2 -4xy-2ay+4ax ]$

$\displaystyle \Rightarrow 45x^2 + 45y^2 - 90ax +45a^2 = 64x^2 + 16y^2 + 16a^2 -64xy-32ay+64ax$

$\displaystyle \Rightarrow 19x^2 -29y^2-64xy+154ax-32ay-29a^2=0$

This is the required equation of the hyperbola.

$\\$

(vi) Given focus is $\displaystyle (2,2),$ directrix is $\displaystyle x + y = 9$ and eccentricity $\displaystyle = 2$

Let $\displaystyle S(2, 2)$ be the focus and $\displaystyle P(x, y)$ be a point on the Hyperbola.

Draw $\displaystyle PM$ perpendicular from $\displaystyle P$ on the directrics. Then, by definition

$\displaystyle SP = e PM$

$\displaystyle \Rightarrow SP^2 = e^2 PM^2$

$\displaystyle \Rightarrow (x-2)^2 + ( y - 2)^2 = (2)^2 \Bigg[ \frac{x+y-9}{\sqrt{(1)^2+(1)^2}} \Bigg]^2 \hspace{2.0cm} [\because e = 2]$

$\displaystyle \Rightarrow x^2 + 4 -4x + y^2 + 4 -4y = \frac{4}{2} \Big[ x+y-9 \Big]^2$

$\displaystyle \Rightarrow x^2 + y^2 -4x - 4y + 8 = 2 [ x+y-9 ]^2$

$\displaystyle \Rightarrow x^2 + y^2 -4x - 4y + 8 = 2[ x^2 + y^2 + 81 +2xy-18y-18x ]$

$\displaystyle \Rightarrow x^2 + y^2 -4x - 4y + 8 = 2x^2+2y^2+162+4xy-36y-36x$

$\displaystyle \Rightarrow x^2 + y^2+4xy-32x-32y+154=0$

This is the required equation of the hyperbola.

$\\$

Question 3: Find the eccentricity, coordinates of the foci, equations of directrices and length of the latus-rectum of the hyperbola

(i) $\displaystyle 9x^2 - 16y^2 = 144$

(ii) $\displaystyle 16x^2 - 9y^2 = -144$

(iii) $\displaystyle 4x^2 - 3y^2 = 36$

(iv) $\displaystyle 3x^2 - y^2 = 4$

(v) $\displaystyle 2x^2 - 3y^2 = 5$

Answer:

(i) Given $\displaystyle 9x^2 - 16y^2 = 144$

$\displaystyle \Rightarrow \frac{9x^2}{144} - \frac{16y^2}{144} = 1$

$\displaystyle \Rightarrow \frac{x^2}{16} - \frac{y^2}{9} = 1$

$\displaystyle \text{This is of the form } \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \text{ where } a^2 = 16 \text{ and } b^2 = 9$

$\displaystyle \text{Eccentricity: } e = \sqrt{1 + \frac{b^2}{a^2} } = \sqrt{ 1+ \frac{9}{16} } = \sqrt{ \frac{25}{16} } = \frac{5}{4}$

$\displaystyle \text{Foci: The coordinates of the foci are } ( \pm ae, 0) \text{ i.e. } (\pm 5, 0)$

$\displaystyle \text{Equation of directrices: } x = \pm \frac{a}{e} \text{ i.e. } x = \pm \frac{16}{5} \hspace{0.2cm} \Rightarrow 5x = \pm 16 \hspace{0.2cm} \Rightarrow 5x \mp 16 = 0$

$\displaystyle \text{Length of latus-rectum } = \frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{9}{2}$

$\\$

(ii) Given $\displaystyle 16x^2 - 9y^2 = -144$

$\displaystyle \Rightarrow \frac{16x^2}{144} - \frac{9y^2}{144} = -1$

$\displaystyle \Rightarrow \frac{x^2}{9} - \frac{y^2}{16} = -1$

$\displaystyle \text{This is of the form } \frac{x^2}{a^2} - \frac{y^2}{b^2} = -1, \text{ where } a^2 = 9 \text{ and } b^2 = 16$

$\displaystyle \text{Eccentricity: } e = \sqrt{1 + \frac{a^2}{b^2} } = \sqrt{ 1+ \frac{9}{16} } = \sqrt{ \frac{25}{16} } = \frac{5}{4}$

$\displaystyle \text{Foci: The coordinates of the foci are } ( 0, \pm be) \text{ i.e. } (0, \pm 5)$

$\displaystyle \text{Equation of directrices: } y = \pm \frac{b}{e} \text{ i.e. } y = \pm \frac{4}{\frac{5}{4}} \hspace{0.2cm} \Rightarrow 5y = \pm 16 \hspace{0.2cm} \Rightarrow 5y \mp 16 = 0$

$\displaystyle \text{Length of latus-rectum } = \frac{2a^2}{b} = \frac{2 \times 9}{4} = \frac{9}{2}$

$\\$

(iii) Given $\displaystyle 4x^2 - 3y^2 = 36$

$\displaystyle \Rightarrow \frac{4x^2}{36} - \frac{3y^2}{36} = 1$

$\displaystyle \Rightarrow \frac{x^2}{9} - \frac{y^2}{12} = 1$

$\displaystyle \text{This is of the form } \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \text{ where } a^2 = 9 \text{ and } b^2 = 12$

$\displaystyle \text{Eccentricity: } e = \sqrt{1 + \frac{b^2}{a^2} } = \sqrt{ 1+ \frac{12}{9} } = \sqrt{ \frac{21}{9} } = \sqrt{ \frac{7}{3} }$

$\displaystyle \text{Foci: The coordinates of the foci are } ( \pm ae, 0) \text{ i.e. } (\pm \sqrt{21} , 0)$

$\displaystyle \text{Equation of directrices: } x = \pm \frac{a}{e} \text{ i.e. } x = \pm 3 \times \frac{\sqrt{3}}{\sqrt{7}} \hspace{0.2cm} \Rightarrow \sqrt{7} x = \pm 3\sqrt{3} \hspace{0.2cm} \Rightarrow \sqrt{7} x \mp 3\sqrt{3} = 0$

$\displaystyle \text{Length of latus-rectum } = \frac{2b^2}{a} = \frac{2 \times 12}{3} = 8$

$\\$

(iv) Given $\displaystyle 3x^2 - y^2 = 4$

$\displaystyle \Rightarrow \frac{3x^2}{4} - \frac{y^2}{4} = 1$

$\displaystyle \Rightarrow \frac{x^2}{\frac{4}{3}} - \frac{y^2}{4} = 1$

$\displaystyle \text{This is of the form } \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \text{ where } a^2 = \frac{4}{3} \text{ and } b^2 = 4$

$\displaystyle \text{Eccentricity: } e = \sqrt{1 + \frac{b^2}{a^2} } = \sqrt{ 1+ \frac{4}{\frac{4}{3}} } = \sqrt{ 1+3 } = 2$

$\displaystyle \text{Foci: The coordinates of the foci are } ( \pm ae, 0) \text{ i.e. } (\pm \frac{4}{\sqrt{3}}, 0)$

$\displaystyle \text{Equation of directrices: } x = \pm \frac{a}{e} \text{ i.e. } x = \pm \frac{\frac{2}{\sqrt{3}}}{2} \hspace{0.2cm} \Rightarrow \sqrt{3}x = \pm 1 \hspace{0.2cm} \\ \\ \Rightarrow \sqrt{3}x \mp 1 = 0$

$\displaystyle \text{Length of latus-rectum } = \frac{2b^2}{a} = \frac{2 \times 4}{\frac{2}{\sqrt{3}}} = 4\sqrt{3}$

$\\$

(v) Given $\displaystyle 2x^2 - 3y^2 = 5$

$\displaystyle \Rightarrow \frac{2x^2}{5} - \frac{3y^2}{5} = 1$

$\displaystyle \Rightarrow \frac{x^2}{\frac{5}{2}} - \frac{y^2}{\frac{5}{3}} = 1$

$\displaystyle \text{This is of the form } \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \text{ where } a^2 = \frac{5}{2} \text{ and } b^2 = \frac{5}{3}$

$\displaystyle \text{Eccentricity: } e = \sqrt{1 + \frac{b^2}{a^2} } = \sqrt{ 1+ \frac{\frac{5}{3}}{\frac{5}{2}} } = \sqrt{ \frac{5}{3} }$

$\displaystyle \text{Foci: The coordinates of the foci are } ( \pm ae, 0) \text{ i.e. } (\pm \frac{5}{\sqrt{6}}, 0)$

$\displaystyle \text{Equation of directrices: } x = \pm \frac{a}{e} \text{ i.e. } x = \pm \frac{\sqrt{\frac{5}{2}}}{\sqrt{\frac{5}{3}}} \hspace{0.2cm} \Rightarrow \sqrt{2}x = \pm \sqrt{3} \hspace{0.2cm} \\ \\ \Rightarrow \sqrt{2}x \mp \sqrt{3} = 0$

$\displaystyle \text{Length of latus-rectum } = \frac{2b^2}{a} = \frac{2 \times \frac{5}{3}}{\sqrt{\frac{5}{2}}} = \frac{10}{3} \sqrt{\frac{2}{5}}$

$\\$

Question 4: Find the axes, eccentricity, latus-rectum and the coordinates of the foci of the hyperbola $\displaystyle 25x^2- 36y^2= 225$

Answer:

Given $\displaystyle 25x^2- 36y^2= 225$

$\displaystyle \Rightarrow \frac{25x^2}{225} - \frac{36y^2}{225} = 1$

$\displaystyle \Rightarrow \frac{x^2}{9} - \frac{4y^2}{25} = 1$

$\displaystyle \Rightarrow \frac{x^2}{9} - \frac{y^2}{\frac{25}{4}} = 1$

$\displaystyle \text{This is of the form } \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \text{ where } a^2 = 9 \text{ and } b^2 = \frac{25}{4}$

$\displaystyle \text{Length of the transverse axis } = 2a = 2 \times 3 = 6$

$\displaystyle \text{Length of conjugate axis } = 2b = 2 \times \frac{5}{2} = 5$

$\displaystyle \text{Eccentricity: } e = \sqrt{1 + \frac{b^2}{a^2} } = \sqrt{ 1+ \frac{ \frac{25}{4} } {9} } = \sqrt{ \frac{61}{36} } = \frac{\sqrt{61}}{6}$

$\displaystyle \text{Length of latus-rectum } = \frac{2b^2}{a} = \frac{2\times \frac{25}{4} }{3} = \frac{25}{6}$

$\displaystyle \text{Foci: The coordinates of the foci are } ( \pm ae, 0) \text{ i.e. } (\pm \frac{\sqrt{61}}{2}, 0)$

$\\$

Question 5: Find the center, eccentricity, foci and directrices of the hyperbola

(i) $\displaystyle 16x^2 - 9y^2 +32x+36y - 164 = 0$

(ii) $\displaystyle x^2 - y^2 + 4x = 0$

(iii) $\displaystyle x^2 - 3y^2 - 2x = 8$

Answer:

(i) Given $\displaystyle 16x^2 - 9y^2 +32x+36y - 164 = 0$ can be simplified in the following way:

$\displaystyle \Rightarrow 16(x^2+2x) - 9(y^2-4y) = 164$

$\displaystyle \Rightarrow 16(x^2+2x+ 1) - 9(y^2-4y+4) = 164+16-36$

$\displaystyle \Rightarrow (x+1)^2 - 9(y-2)^2 = 144$

$\displaystyle \Rightarrow \frac{(x+1)^2}{9} - \frac{(y-2)^2}{16} = 1 \text{ ... ... ... ... ... i)}$

Shifting the origin at $\displaystyle ( -1, 2)$ without rotating the coordinate axes and denoting the new coordinates with respect to the new axes $\displaystyle X$ and $\displaystyle Y,$

$\displaystyle \text{We have } X = x+1 \text{ and } Y = y-2 \text{ ... ... ... ... ... ii)}$

$\displaystyle \text{This is of the form } \frac{(X)^2}{a^2} - \frac{(Y)^2}{b^2} = 1 \text{ where } a^2 = 9 \text{ and } b^2 = 16$

Center: w.r.t the new axes $\displaystyle (X=0, Y=0)$

$\displaystyle \therefore x = -1 \text{ and } y = 2$

Therefore the coordinates of the center with respect to old axes are $\displaystyle (-1, 2)$

$\displaystyle \text{Eccentricity: } e = \sqrt{1 + \frac{b^2}{a^2} } = \sqrt{ 1+ \frac{ 16 }{9} } = \sqrt{ \frac{25}{9} } = \frac{5}{3}$

$\displaystyle \text{Foci: The coordinates of the foci are with respect to the new axes is given by } \\ \\ ( X=\pm ae, Y=0) \text{ i.e. } (X=\pm 5, Y=0)$

$\displaystyle \text{Putting } X = \pm 5 \text{ and } Y = 0 \text{in equation ii) we get }$

$\displaystyle x = \pm 5 - 1 \text{ and } y = 0 +2$

$\displaystyle \Rightarrow x = 4, -6 \text{ and } y = 2$

$\displaystyle \text{Therefore the foci are } (4, 2) \text{ and } ( -6, 2)$

Equation of directrix :

$\displaystyle X = \pm \frac{a}{e} = \pm \frac{3}{\frac{5}{3}} = \pm \frac{9}{5}$

$\displaystyle \text{Putting } X = \pm \frac{9}{5} \text{ in equation ii) we get }$

$\displaystyle \Rightarrow x = \pm \frac{9}{5} - 1$

$\displaystyle \Rightarrow x = \frac{4}{5} \text{ and } x = \frac{-14}{5}$

$\displaystyle \Rightarrow 5x - 4 = 0 \text{ and } 5x + 14 = 0$

Hence the equations of the directrices w.r.t the old axes are $\displaystyle 5x - 4 = 0$ and $\displaystyle 5x + 14 = 0$

$\\$

(ii) Given $\displaystyle x^2 - y^2 + 4x = 0$ can be simplified in the following way:

$\displaystyle \Rightarrow x^2+4x-y^2=0$

$\displaystyle \Rightarrow x^2+4x+4-y^2=4$

$\displaystyle \Rightarrow (x+2)^2-y^2=4$

$\displaystyle \Rightarrow \frac{(x+2)^2}{4} - \frac{(y)^2}{4} = 1 \text{ ... ... ... ... ... i)}$

Shifting the origin at $\displaystyle ( -2, 0)$ without rotating the coordinate axes and denoting the new coordinates with respect to the new axes $\displaystyle X$ and $\displaystyle Y,$

$\displaystyle \text{We have } X = x+2 \text{ and } Y = y \text{ ... ... ... ... ... ii)}$

$\displaystyle \text{This is of the form } \frac{(X)^2}{a^2} - \frac{(Y)^2}{b^2} = 1 \text{ where } a^2 =4 \text{ and } b^2 = 4$

Center: w.r.t the new axes $\displaystyle (X=0, Y=0)$

$\displaystyle \therefore x = -2 \text{ and } y = 0$

Therefore the coordinates of the center with respect to old axes are $\displaystyle (-2, 0)$

$\displaystyle \text{Eccentricity: } e = \sqrt{1 + \frac{b^2}{a^2} } = \sqrt{ 1+ \frac{ 4 }{4} } = \sqrt{ 2 }$

$\displaystyle \text{Foci: The coordinates of the foci are with respect to the new axes is given by } \\ \\ ( X=\pm ae, Y=0) \text{ i.e. } (X=\pm 2\sqrt{2}, Y=0)$

$\displaystyle \text{Putting } X = \pm 2\sqrt{2} \text{ and } Y = 0 \text{in equation ii) we get }$

$\displaystyle x = \pm 2\sqrt{2} - 2 \text{ and } y = 0$

$\displaystyle \text{Therefore the foci are } (-2\pm 2\sqrt{2}, 0)$

Equation of directrix :

$\displaystyle X = \pm \frac{a}{e} = \pm \frac{2}{\sqrt{2}}$

$\displaystyle \text{Putting } X = \pm \frac{2}{\sqrt{2}} \text{ in equation ii) we get }$

$\displaystyle \Rightarrow X = \pm \frac{2}{\sqrt{2}} - 2$

$\displaystyle \Rightarrow x +2= \pm \sqrt{2}$

Hence the equations of the directrices w.r.t the old axes are $\displaystyle x +2= \pm \sqrt{2}$

$\\$

(iii) Given $\displaystyle x^2 - 3y^2 - 2x = 8$ can be simplified in the following way:

$\displaystyle \Rightarrow x^2-2x-3y=8$

$\displaystyle \Rightarrow x^2-2x+1-1-3y^2=8$

$\displaystyle \Rightarrow (x-1)^2-3y^2=9$

$\displaystyle \Rightarrow \frac{(x-1)^2}{9} - \frac{(y)^2}{3} = 1 \text{ ... ... ... ... ... i)}$

Shifting the origin at $\displaystyle ( 1, 0)$ without rotating the coordinate axes and denoting the new coordinates with respect to the new axes $\displaystyle X$ and $\displaystyle Y,$

$\displaystyle \text{We have } X = x-1 \text{ and } Y = y \text{ ... ... ... ... ... ii)}$

$\displaystyle \text{This is of the form } \frac{(X)^2}{a^2} - \frac{(Y)^2}{b^2} = 1 \text{ where } a^2 =9 \text{ and } b^2 = 3$

Center: w.r.t the new axes $\displaystyle (X=0, Y=0)$

$\displaystyle \therefore x = 1 \text{ and } y = 0$

Therefore the coordinates of the center with respect to old axes are $\displaystyle (1, 0)$

$\displaystyle \text{Eccentricity: } e = \sqrt{1 + \frac{b^2}{a^2} } = \sqrt{ 1+ \frac{ 3 }{9} } = \sqrt{ \frac{4}{3} } = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

$\displaystyle \text{Foci: The coordinates of the foci are with respect to the new axes is given by } \\ \\ ( X=\pm ae, Y=0) \text{ i.e. } (X=\pm 2\sqrt{3}, Y=0)$

$\displaystyle \text{Putting } X = \pm 2\sqrt{3} \text{ and } Y = 0 \text{in equation ii) we get }$

$\displaystyle x = \pm 2\sqrt{3} +1 \text{ and } y = 0$

$\displaystyle \text{Therefore the foci are } (1\pm 2\sqrt{3}, 0)$

Equation of directrix :

$\displaystyle X = \pm \frac{a}{e} = \pm \frac{3}{\frac{2\sqrt{3}}{3}} = \pm \frac{9}{2\sqrt{3}}$

$\displaystyle \text{Putting } X = \pm \frac{9}{2\sqrt{3}} \text{ in equation ii) we get }$

$\displaystyle \Rightarrow x = 1 \pm \frac{9}{2\sqrt{3}}$

Hence the equations of the directrices w.r.t the old axes are $\displaystyle x = 1 \pm \frac{9}{2\sqrt{3}}$

$\\$

Question 6: Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in

the following cases:

(i) the distance between the foci $\displaystyle = 16$ and eccentricity $\displaystyle = \sqrt{2}$

(ii) conjugate axis is $\displaystyle 5$ and the distance between foci $\displaystyle = 13$

(iii) conjugate axis is $\displaystyle 7$ and passes through the point $\displaystyle (3, - 2).$

Answer:

(i) Given the distance between the foci $\displaystyle = 16$ and eccentricity $\displaystyle = \sqrt{2}$

The distance between the foci is $\displaystyle 2ae$

$\displaystyle \text{Given } 2ae = 16 \Rightarrow ae = 8$

$\displaystyle \text{Given } e = \sqrt{2}$

$\displaystyle \therefore a(\sqrt{2}) = 8 \Rightarrow a = 4 \sqrt{2}$

$\displaystyle \therefore a^2 = 32$

$\displaystyle \text{We know, } b^2 = a^2(e^2-1) = 32(2-1) = 32$

Therefore the equation of the hyperbola is given as

$\displaystyle \frac{x^2}{32} - \frac{y^2}{32} = 1$

$\displaystyle \Rightarrow x^2 - y^2 = 31$

$\\$

(ii) Given conjugate axis is $\displaystyle 5$ and the distance between foci $\displaystyle = 13$

The distance between the foci is $\displaystyle 2ae$

$\displaystyle \text{Given } 2ae = 13 \Rightarrow ae = \frac{13}{2}$

Length of conjugate axis,

$\displaystyle 2b = 5 \Rightarrow b = \frac{5}{2}$

$\displaystyle \text{We know, } b^2 = a^2(e^2-1)$

$\displaystyle \Rightarrow \Big(\frac{5}{2} \Big)^2 = \Big(\frac{13}{2} \Big)^2 - a^2$

$\displaystyle \Rightarrow a^2 = \frac{169-25}{4} = \frac{144}{4} = 36$

$\displaystyle \Rightarrow a = 6$

Therefore the equation of the hyperbola is given as

$\displaystyle \frac{x^2}{36} - \frac{4y^2}{25} = 1$

$\displaystyle \Rightarrow 25x^2 -144 y^2 = 900$

$\\$

(iii) Given conjugate axis is $\displaystyle 7$ and passes through the point $\displaystyle (3, - 2).$

Length of the conjugate axis

$\displaystyle 2b = 7 \Rightarrow b = \frac{7}{2}$

$\displaystyle \text{Let the equation of the hyperbola be } \frac{x^2}{a^2} - \frac{4y^2}{b^2} = 1$

It passes through $\displaystyle ( 3, -2)$

$\displaystyle \therefore \frac{3^2}{a^2} - \frac{(-2)^2}{(\frac{7}{2})^2} = 1$

$\displaystyle \Rightarrow \frac{3^2}{a^2} - \frac{16}{49} = 1$

$\displaystyle \Rightarrow \frac{9}{a^2} = \frac{16}{49} + 1$

$\displaystyle \Rightarrow \frac{9}{a^2} = \frac{65}{49}$

$\displaystyle \Rightarrow a^2 = \frac{441}{65}$

Therefore the equation of the hyperbola is given as

$\displaystyle \frac{65x^2}{441} - \frac{4y^2}{49} = 1$

$\displaystyle \Rightarrow 65x^2 -36 y^2 = 441$

$\\$

Question 7: Find the equation of the hyperbola whose

(i) foci are $\displaystyle (6,4)$ and $\displaystyle (- 4,4)$ and eccentricity is $\displaystyle 2.$

(ii) vertices are $\displaystyle (- 8, - 1)$ and $\displaystyle (16,- 1)$ and focus is $\displaystyle (17,- 1)$

(iii) foci are $\displaystyle (4,2)$ and $\displaystyle (8,2)$ and eccentricity is $\displaystyle 2.$

(iv) vertices are at $\displaystyle (0 , \pm 7)$ and foci are $\displaystyle ( 0, \pm \frac{28}{3} )$

(v) vertices are at $\displaystyle (\pm 6,0)$ and one of the directrices is $\displaystyle x = 4$

(vi) foci at $\displaystyle (\pm 2,0 )$ and eccentricity is $\displaystyle 3/2$

Answer:

(i) Given foci are $\displaystyle (6,4)$ and $\displaystyle (- 4,4)$ and eccentricity is $\displaystyle 2.$

The center of the hyperbola is the mid point of the line joining the two foci.

$\displaystyle \text{Therefore the coordinates of the center are } (\frac{6-4}{2}, \frac{4+4}{2}) \text{ i.e. } (1, 4)$

Let $\displaystyle 2a$ and $\displaystyle 2b$ be the length of transverse and conjugate axes and let $\displaystyle e$ be the eccentricity

Then, the equation of the hyperbola is

$\displaystyle \frac{(x-1)^2}{a^2} - \frac{(y-4)^2}{b^2} = 1 \ \ \ \text{ ... ... ... ... ... i)}$

Now, the distance between the two foci $\displaystyle = 2ae$

$\displaystyle \Rightarrow \sqrt{(6+4)^2+(4-4)^2} = 2ae$

$\displaystyle \Rightarrow 10 = 2ae$

$\displaystyle \Rightarrow 2 \times a \times 2 = 10 \hspace{2.0cm} [ \because e = 2]$

$\displaystyle \Rightarrow a = \frac{5}{2}$

$\displaystyle \Rightarrow a^2 = \frac{25}{4}$

$\displaystyle \text{Now, } b^2 = a^2 ( e^2 -1)$

$\displaystyle \Rightarrow b^2 = \frac{25}{4} ( 4 - 1 )$

$\displaystyle \Rightarrow b^2 = \frac{75}{4}$

$\displaystyle \text{Substituting } a^2 = \frac{25}{4} \text{ and } b^2 = \frac{75}{4} \text{in equation i) we get }$

$\displaystyle \frac{(x-1)^2}{\frac{25}{4}} - \frac{(y-4)^2}{\frac{75}{4}} = 1$

$\displaystyle \Rightarrow \frac{4(x-1)^2}{25} - \frac{4(y-4)^2}{75} = 1$

$\displaystyle \Rightarrow \frac{12(x-1)^2-4(y-4)^2}{75} = 1$

$\displaystyle \Rightarrow 12(x-1)^2 - 4 ( y-4)^2 = 75$

$\displaystyle \Rightarrow 12 ( x^2 + 1 - 2x) - 4 ( y^2 + 16 - 8 x) = 75$

$\displaystyle \Rightarrow 12x^2 + 12 - 24 x - 4y^2 - 64 +32y = 75$

$\displaystyle \Rightarrow 12x^2 - 4y^2 - 24x + 32 y - 127 = 0$

This is the equation of the required hyperbola.

$\\$

(ii) Given vertices are $\displaystyle (- 8, - 1)$ and $\displaystyle (16,- 1)$ and focus is $\displaystyle (17,- 1)$

The center of the hyperbola is the mid point of the line joining the two foci.

$\displaystyle \text{Therefore the coordinates of the center are } (\frac{16-8}{2}, \frac{-1-1}{2}) \text{ i.e. } (4, -1)$

Let $\displaystyle 2a$ and $\displaystyle 2b$ be the length of transverse and conjugate axes and let $\displaystyle e$ be the eccentricity

Then, the equation of the hyperbola is

$\displaystyle \frac{(x-1)^2}{a^2} - \frac{(y-4)^2}{b^2} = 1 \ \ \ \text{ ... ... ... ... ... i)}$

Now, the distance between the two vertices $\displaystyle = 2a$

$\displaystyle \Rightarrow \sqrt{(16+8)^2+(-1+1)^2} = 2a$

$\displaystyle \Rightarrow 24 = 2a$

$\displaystyle \Rightarrow a =12$

$\displaystyle \Rightarrow a^2 = 144$

The distance between the focus and vertex $\displaystyle = ae-a$

$\displaystyle \therefore \sqrt{(17-16)^2+(-1+1)^2} = ae-a$

$\displaystyle \Rightarrow \sqrt{1^1} = ae-a$

$\displaystyle \Rightarrow ae-a = 1$

$\displaystyle \Rightarrow 12 e = 1 + 12 \hspace{2.0cm} [ \because a = 12]$

$\displaystyle \Rightarrow e = \frac{13}{12}$

$\displaystyle \Rightarrow e^2 = \frac{169}{144}$

$\displaystyle \text{Now, } b^2 = a^2 ( e^2 -1)$

$\displaystyle \Rightarrow b^2 = 144 (\frac{169}{144} - 1 )$

$\displaystyle \Rightarrow b^2 = 25$

$\displaystyle \text{Substituting } a^2 = 144\text{ and } b^2 = 25 \text{in equation i) we get }$

$\displaystyle \frac{(x-4)^2}{144} - \frac{(y=1)^2}{25} = 1$

$\displaystyle \Rightarrow \frac{25(x-4)^2-144(y+1)^2}{3600} = 1$

$\displaystyle \Rightarrow 25(x-4)^2 - 144 ( y+1)^2 = 3600$

$\displaystyle \Rightarrow 25 ( x^2 + 16 - 8x) - 144 ( y^2 + 1 +2 x) = 3600$

$\displaystyle \Rightarrow 25x^2 + 400 - 200 x - 144y^2 - 144 -288y = 3600$

$\displaystyle \Rightarrow 25x^2 - 144y^2 - 200x -288 y - 3344 = 0$

This is the equation of the required hyperbola.

$\\$

(iii) Given foci are $\displaystyle (4,2)$ and $\displaystyle (8,2)$ and eccentricity is $\displaystyle 2.$

The center of the hyperbola is the mid point of the line joining the two foci.

$\displaystyle \text{Therefore the coordinates of the center are } (\frac{4+8}{2}, \frac{2+2}{2}) \text{ i.e. } (6, 2)$

Let $\displaystyle 2a$ and $\displaystyle 2b$ be the length of transverse and conjugate axes and let $\displaystyle e$ be the eccentricity

Then, the equation of the hyperbola is

$\displaystyle \frac{(x-1)^2}{a^2} - \frac{(y-4)^2}{b^2} = 1 \ \ \ \text{ ... ... ... ... ... i)}$

Now, the distance between the two foci $\displaystyle = 2ae$

$\displaystyle \Rightarrow \sqrt{(8-4)^2+(2-2)^2} = 2ae$

$\displaystyle \Rightarrow 4 = 2ae$

$\displaystyle \Rightarrow 2 \times a \times 2 = 4 \hspace{2.0cm} [ \because e = 2]$

$\displaystyle \Rightarrow a = 1$

$\displaystyle \Rightarrow a^2 = 1$

$\displaystyle \text{Now, } b^2 = a^2 ( e^2 -1)$

$\displaystyle \Rightarrow b^2 = 1 ( 2^2 - 1 ) \hspace{2.0cm} [ \because e = 2]$

$\displaystyle \Rightarrow b^2 = 3$

$\displaystyle \text{Substituting } a^2 = 1 \text{ and } b^2 = 3 \text{in equation i) we get }$

$\displaystyle \frac{(x-6)^2}{1} - \frac{(y-2)^2}{3} = 1$

$\displaystyle \Rightarrow \frac{3(x-6)^2-(y-2)^2}{3} = 1$

$\displaystyle \Rightarrow 3(x-6)^2-(y-2)^2 = 3$

$\displaystyle \Rightarrow 3 ( x^2 +36 - 12x) - ( y^2 + 4 - 4 x) = 3$

$\displaystyle \Rightarrow 3x^2 + 108 - 36 x - y^2 -4 +4y = 3$

$\displaystyle \Rightarrow 3x^2 - y^2 - 36x + 4 y +101 = 0$

This is the equation of the required hyperbola.

$\\$

(iv) Given vertices are at $\displaystyle (0 , \pm 7)$ and foci are $\displaystyle ( 0, \pm \frac{28}{3} )$

Since the vertices are on y-axis, the equation of the required hyperbola is

$\displaystyle \frac{(x-1)^2}{a^2} - \frac{(y-4)^2}{b^2} = -1 \ \ \ \text{ ... ... ... ... ... i)}$

The coordinates of its vertices and foci are $\displaystyle (0, \pm b)$ and $\displaystyle (0, \pm be)$ respectively.

Therefore $\displaystyle b = 7$ and $\displaystyle b^2 = 49$

Now, $\displaystyle be = \frac{28}{3}$

$\displaystyle \Rightarrow 7 \times e = \frac{28}{3}$

$\displaystyle \Rightarrow e = \frac{4}{3}$

$\displaystyle \Rightarrow e^2 = \frac{16}{9}$

We know, $\displaystyle a^2 = b^2 ( e^2 - 1)$

$\displaystyle \Rightarrow a^2 = 49 9 \frac{16}{9} - 1)$

$\displaystyle \Rightarrow a^2 = 49 \times \frac{7}{9}$

$\displaystyle \Rightarrow a^2 = \frac{343}{9}$

$\displaystyle \text{Substituting } a^2 = \frac{343}{9} \text{ and } b^2 = 49 \text{in equation i) we get }$

$\displaystyle \frac{x^2}{\frac{343}{9}} - \frac{y^2}{49} = -1$

$\displaystyle \Rightarrow \frac{9x^2}{343} - \frac{y^2}{49} = -1$

This is the equation of the required hyperbola.

$\\$

(v) Given vertices are at $\displaystyle (\pm 6,0)$ and one of the directrices is $\displaystyle x = 4$

The vertices of the hyperbola is $\displaystyle (\pm a, 0).$

Therefore $\displaystyle a = 6$ and $\displaystyle a^2 = 36$

Given $\displaystyle x = 4$

$\displaystyle \Rightarrow \frac{a}{e} = 4$

$\displaystyle \Rightarrow e = \frac{6}{4} = \frac{3}{2}$

We know $\displaystyle b^2 = a^2 ( e^2 -1)$

$\displaystyle \Rightarrow b^2 = 36( \frac{9}{4} - 1) = 45$

$\displaystyle \text{Substituting } a^2 = \frac{343}{9} \text{ and } b^2 = 49$

$\displaystyle \frac{x^2}{36} - \frac{y^2}{45} = 1$

This is the equation of the required hyperbola.

$\\$

(vi) Given foci at $\displaystyle (\pm 2,0 )$ and eccentricity is $\displaystyle 3/2$

The foci of the hyperbola is $\displaystyle (\pm ae, 0).$

Therefore $\displaystyle ae = 2$

$\displaystyle \Rightarrow a \times \frac{3}{2} = 2$

$\displaystyle \Rightarrow a = \frac{4}{3}$

$\displaystyle \Rightarrow a^2 = \frac{16}{9}$

We know $\displaystyle b^2 = a^2 ( e^2 -1)$

$\displaystyle \Rightarrow b^2 = \frac{16}{9} ( \frac{9}{4} - 1) = \frac{20}{9}$

$\displaystyle \text{Substituting } a^2 = \frac{16}{9} \text{ and } b^2 = \frac{20}{9}$

$\displaystyle \frac{x^2}{\frac{16}{9}} - \frac{y^2}{\frac{20}{9}} = 1$

$\displaystyle \Rightarrow \frac{x^2}{4} - \frac{y^2}{5} = \frac{4}{9}$

This is the equation of the required hyperbola.

$\\$

Question 8: Find the eccentricity of the hyperbola, the length of whose conjugate axis is $\displaystyle \frac{3}{4}$ of the length of transverse axis.

Answer:

Let $\displaystyle 2a$ and $\displaystyle 2b$ be the transverse and conjugate axes and e be the eccentricity. Then,

$\displaystyle \text{The length of the conjugate axis } = \frac{3}{4} [ \text{ length of the transverse axis } ]$

$\displaystyle \Rightarrow 2b = \frac{3}{4} ( 2a)$

$\displaystyle \Rightarrow \frac{b}{a} = \frac{3}{4}$

$\displaystyle \Rightarrow \frac{b^2}{a^2} = \frac{9}{16}$

$\displaystyle \text{Now, } e = \sqrt{1 +\frac{b^2}{a^2} } = \sqrt{1 +\frac{9}{16} } = \sqrt{\frac{25}{16}} = \frac{5}{4}$

$\displaystyle \text{Hence } e = \frac{5}{4}$

$\\$

Question 9: Find the equation of the hyperbola whose

(i) focus is at $\displaystyle (5,2),$ vertex at $\displaystyle (4,2)$ and center at $\displaystyle (3,2)$

(ii) focus is at $\displaystyle (4,2),$ center at $\displaystyle (6,2)$ and $\displaystyle e = 2.$

Answer:

(i) Given focus is at $\displaystyle (5,2),$ vertex at $\displaystyle (4,2)$ and center at $\displaystyle (3,2)$

Let $\displaystyle (x_1, y_1)$ be the coordinate of the second vertex.

We know that the vertex of the hyperbola is the mid point of the line joining the two vertices.

$\displaystyle \therefore \frac{x_1+4}{2}= 3 \text{ and } \frac{y_1+2}{2}= 2$

$\displaystyle \Rightarrow x_1 = 2 \text{ and } y_1 = 2$

Therefore the coordinates of the second vertex is $\displaystyle (2, 2 )$

Let $\displaystyle 2a$ and $\displaystyle 2b$ be the length of transverse and conjugate axes and let $\displaystyle e$ be the eccentricity. The the equation of the hyperbola is

$\displaystyle \frac{(x-3)^3}{a^2} - \frac{(y-2)^2}{b^2} = 1 \ \ \ \text{ ... ... ... ... ... i) }$

Now, the distance between the two vertices $\displaystyle = 2a$

$\displaystyle \Rightarrow \sqrt{ (4-2)^2 + ( 2-2)^2} = 2a$

$\displaystyle \Rightarrow \sqrt{2^2} = 2a$

$\displaystyle \Rightarrow 2a = 2$

$\displaystyle \Rightarrow a = 1$

Now, the distance between the vertex and focus is $\displaystyle = ae - a$

$\displaystyle \Rightarrow \sqrt{ (5-4)^2 + ( 2-2)^2} = ae - a$

$\displaystyle \Rightarrow \sqrt{1^2} = ae - a$

$\displaystyle \Rightarrow 1 \times e - 1 = 1$

$\displaystyle \Rightarrow e = 2$

Now, $\displaystyle b^2 = a^2 ( e^2 -1) = 1^2 ( 2^2 -1) = 3$

Substituting $\displaystyle a^2 = 1$ and $\displaystyle ^2 = 3$ in equation i) we get

$\displaystyle \frac{(x-3)^3}{1} - \frac{(y-2)^2}{3} = 1$

$\displaystyle \Rightarrow \frac{3(x-3)^2-(y-2)^2}{3} = 1$

$\displaystyle \Rightarrow 3(x-3)^2 - (y-2)^2 = 3$

This is the equation of the required hyperbola.

$\\$

(ii) Given focus is at $\displaystyle (4,2),$ center at $\displaystyle (6,2)$ and $\displaystyle e = 2$

Let $\displaystyle (x_1, y_1)$ be the coordinate of the second focus of the hyperbola.

We know that the vertex of the hyperbola is the mid point of the line joining the two foci.

$\displaystyle \therefore \frac{x_1+4}{2}= 6 \text{ and } \frac{y_1+2}{2}= 2$

$\displaystyle \Rightarrow x_1 = 8 \text{ and } y_1 = 2$

Therefore the coordinates of the second focus is $\displaystyle (8, 2 )$

Let $\displaystyle 2a$ and $\displaystyle 2b$ be the length of transverse and conjugate axes and let $\displaystyle e$ be the eccentricity. The the equation of the hyperbola is

$\displaystyle \frac{(x-6)^3}{a^2} - \frac{(y-2)^2}{b^2} = 1 \ \ \ \text{ ... ... ... ... ... i) }$

Now, the distance between the two vertices $\displaystyle = 2a$

$\displaystyle \Rightarrow \sqrt{ (8-4)^2 + ( 2-2)^2} = 2a$

$\displaystyle \Rightarrow \sqrt{2^2} = 2a$

$\displaystyle \Rightarrow 2a = 2$

$\displaystyle \Rightarrow a = 1$

Given $\displaystyle e = 2$

Now, $\displaystyle b^2 = a^2 ( e^2 -1) = 1^2 ( 2^2 -1) = 3$

Substituting $\displaystyle a^2 = 1$ and $\displaystyle ^2 = 3$ in equation i) we get

$\displaystyle \frac{(x-6)^3}{1} - \frac{(y-2)^2}{3} = 1$

$\displaystyle \Rightarrow \frac{3(x-6)^2-(y-2)^2}{3} = 1$

$\displaystyle \Rightarrow 3(x-6)^2 - (y-2)^2 = 3$

This is the equation of the required hyperbola.

$\\$

Question 10: If $\displaystyle P$ is any point on the hyperbola whose axis are equal, prove that $\displaystyle SP.SP = CP^2$

Answer:

$\displaystyle \text{Equation of the hyperbola is } \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

If the axes of the hyperbola are equal, then $\displaystyle a = b$

The, the equation of hyperbola becomes $\displaystyle x^2 - y^2 = a^2$

$\displaystyle \therefore b^2 = a^2 ( e^2 - 1)$

$\displaystyle \Rightarrow a^2 = a^2 ( e^2 -1 )$

$\displaystyle \Rightarrow 1 = ( e^2 -1 )$

$\displaystyle \Rightarrow e^2 = 2$

$\displaystyle \Rightarrow e = \sqrt{2}$

Thus the center $\displaystyle C(0,0)$ and the focus are given by $\displaystyle S(\sqrt{2} a , 0)$ and $\displaystyle S'(-\sqrt{2}a, 0)$ respectively

Let $\displaystyle P ( \alpha, \beta)$ be any point on hyperbola. Therefore it will satisfy the equation. We get,

$\displaystyle \alpha^2 - \beta^2 = a^2$

$\displaystyle \therefore SP^2 = ( \sqrt{2}a - \alpha)^2 + \beta^2$

$\displaystyle \Rightarrow SP^2 = 2a^2 + \alpha^2 - 2 \sqrt{2} a \alpha + \beta^2$

$\displaystyle S'P^2= ( -\sqrt{2} a - \alpha)^2 + \beta^2$

$\displaystyle \Rightarrow S'P^2 = 2a^2 + \alpha^2 + 2 \sqrt{2} a \alpha + \beta^2$

Now, $\displaystyle SP^2 \cdot S'P^2 = (2a^2 + \alpha^2 - 2 \sqrt{2} a \alpha + \beta^2)(2a^2 + \alpha^2 + 2 \sqrt{2} a \alpha + \beta^2)$

$\displaystyle = 4a^2 + 4a^2 (\alpha^2 + \beta^2)+ (\alpha^2 + \beta^2)^2 - 8a^2\alpha^2$

$\displaystyle = 4a^2 ( a^2 - 2\alpha^2) + 4a^2 (\alpha^2 + \beta^2) + (\alpha^2 + \beta^2)^2$

$\displaystyle = 4a^2 ( \alpha^2 - \beta^2 - 2 \alpha^2) + 4a^2(\alpha^2 + \beta^2) + (\alpha^2 + \beta^2)^2$

$\displaystyle = -4a^2 (\alpha^2 + \beta^2)+ 4a^2(\alpha^2 + \beta^2) + (\alpha^2 + \beta^2)^2$

$\displaystyle = (\alpha^2 + \beta^2)^2$

$\displaystyle = CP^4$

$\displaystyle \therefore SP \cdot S'P = CP^2$

$\\$

Question 11: In each of the following find the equations of the hyperbola satisfying the given conditions:

(i) vertices $\displaystyle (\pm 2, 0),$ foci $\displaystyle (\pm 3, 0)$

(ii) vertices $\displaystyle (0, \pm 5),$ foci $\displaystyle (0, \pm 8)$

(iii) vertices $\displaystyle (0, \pm 3),$ foci $\displaystyle (0, \pm 5)$

(iv) foci $\displaystyle (\pm 5, 0),$ transverse axis $\displaystyle = 8$

(v) foci $\displaystyle (0, \pm 13),$ conjugate axis $\displaystyle =24$

(vi) foci $\displaystyle (\pm 3\sqrt{5}, 0)$ the latus-rectum $\displaystyle = 8$

(vii) foci $\displaystyle (\pm 4, 0),$ the latus-rectum $\displaystyle = 12$

(viii) vertices $\displaystyle (0, \pm 6) , e = \frac{5}{3}$

(ix) foci $\displaystyle ( 0, \pm \sqrt{10}),$ passing through $\displaystyle (2, 3)$

(x) foci $\displaystyle ( 0, \pm \sqrt{12}),$ latus-rectum $\displaystyle = 36$

Answer:

(i) Given vertices $\displaystyle (\pm 2, 0),$ foci $\displaystyle (\pm 3, 0)$

$\displaystyle \text{Let the equation of the hyperbola be } \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \text{ ... ... ... ... ... i)}$

The coordinates of its vertices and foci are $\displaystyle ( \pm a, 0)$ and $\displaystyle ( \pm ae, 0)$ respectively.

$\displaystyle \therefore a = 2 \text{ and } a^2 = 4$

$\displaystyle \text{Also } ae = 3 \Rightarrow e = \frac{3}{2}$

$\displaystyle \text{Now, } b^2 = a^2(e^2 -1) = 2^2 \Big( \Big(\frac{3}{2} \Big)^2 - 1 \Big) = 4 \Big( \frac{9-4}{4} \Big) = 4 \times \frac{5}{4} = 5$

Substituting $\displaystyle a^2 = 4$ and $\displaystyle b^2 = 5$ in equation i) we get the equation of the required hyperbola

$\displaystyle \frac{x^2}{4} - \frac{y^2}{5} = 1$

This is the equation of the required hyperbola.

$\\$

(ii) Given vertices $\displaystyle (0, \pm 5),$ foci $\displaystyle (0, \pm 8)$

Since the vertices lie on y-axis, so let the equation of the required hyperbola be

$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = -1 \text{ ... ... ... ... ... i)}$

The coordinates of its vertices and foci are $\displaystyle ( 0, \pm b)$ and $\displaystyle ( 0, \pm ae)$ respectively.

$\displaystyle \therefore b = 5 \text{ and } b^2 = 25$

$\displaystyle \text{Also } be = 8 \Rightarrow e = \frac{8}{5}$

$\displaystyle \text{Now, } a^2 = b^2(e^2 -1) = 25 \Big( \Big(\frac{8}{5} \Big)^2 - 1 \Big) = 25 \Big( \frac{39}{25} \Big) = 39$

Substituting $\displaystyle a^2 = 39$ and $\displaystyle b^2 = 25$ in equation i) we get the equation of the required hyperbola

$\displaystyle \frac{x^2}{39} - \frac{y^2}{25} = -1$

This is the equation of the required hyperbola.

$\\$

(iii) Given vertices $\displaystyle (0, \pm 3),$ foci $\displaystyle (0, \pm 5)$

Since the vertices lie on y-axis, so let the equation of the required hyperbola be

$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = -1 \text{ ... ... ... ... ... i)}$

The coordinates of its vertices and foci are $\displaystyle ( 0, \pm b)$ and $\displaystyle ( 0, \pm ae)$ respectively.

$\displaystyle \therefore b = 3 \text{ and } b^2 = 9$

$\displaystyle \text{Also } be = 5 \Rightarrow e = \frac{5}{3}$

$\displaystyle \text{Now, } a^2 = b^2(e^2 -1) = 9 \Big( \Big(\frac{5}{3} \Big)^2 - 1 \Big) = 9 \Big( \frac{16}{9} \Big) = 16$

Substituting $\displaystyle a^2 = 16$ and $\displaystyle b^2 = 9$ in equation i) we get the equation of the required hyperbola

$\displaystyle \frac{x^2}{16} - \frac{y^2}{9} = -1$

This is the equation of the required hyperbola.

$\\$

(iv) Given foci $\displaystyle (\pm 5, 0),$ transverse axis $\displaystyle = 8$

$\displaystyle \text{Let the equation of the hyperbola be } \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \text{ ... ... ... ... ... i)}$

The coordinates of its vertices and foci are $\displaystyle ( \pm a, 0)$ and $\displaystyle ( \pm ae, 0)$ respectively.

Length of transverse axis $= 8$

$\displaystyle \therefore 2a = 8 \Rightarrow a = 4 \text{ and } a^2 = 16$

$\displaystyle \text{Also } ae = 5 \Rightarrow e = \frac{5}{4}$

$\displaystyle \text{Now, } b^2 = a^2(e^2 -1) = 4^2 \Big( \Big(\frac{5}{4} \Big)^2 - 1 \Big) = 16 \Big( \frac{25-16}{16} \Big) = 16 \times \frac{9}{16} = 9$

Substituting $\displaystyle a^2 = 16$ and $\displaystyle b^2 = 9$ in equation i) we get the equation of the required hyperbola

$\displaystyle \frac{x^2}{16} - \frac{y^2}{9} = 1$

This is the equation of the required hyperbola.

$\\$

(v) Given foci $\displaystyle (0, \pm 13),$ conjugate axis $\displaystyle =24$

Since the vertices lie on x-axis, so let the equation of the required hyperbola be

$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = -1 \text{ ... ... ... ... ... i)}$

The length of the conjugate axis of the required hyperbola is 24

$\displaystyle \therefore 2a = 25 \Rightarrow a = 12 \text{ and } a^2 = 144$

The coordinates of foci of the required hyperbola is (0, \pm be)

$\displaystyle \text{Also } be = 13$

$\displaystyle \text{Now, } a^2 = b^2(e^2 -1) \Rightarrow 144 = 169 - b^2 \Rightarrow b^2 = 25$

Substituting $\displaystyle a^2 = 144$ and $\displaystyle b^2 = 25$ in equation i) we get the equation of the required hyperbola

$\displaystyle \frac{x^2}{144} - \frac{y^2}{25} = -1$

This is the equation of the required hyperbola.

$\\$

(vi) Given foci $\displaystyle (\pm 3\sqrt{5}, 0)$ the latus-rectum $\displaystyle = 8$

Since the vertices lie on x-axis, so let the equation of the required hyperbola be

$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \text{ ... ... ... ... ... i)}$

The length of the conjugate axis of the required hyperbola is $\displaystyle 8.$

$\displaystyle \therefore \frac{2b^2}{a} = 8$

$\displaystyle \Rightarrow b^2 = 4a \text{ ... ... ... ... ... ii)}$

Now, the coordinates of foci of the required hyperbola is $\displaystyle ( \pm ae , 0)$

$\displaystyle \therefore ae = 3\sqrt{5}$

$\displaystyle \Rightarrow e = \frac{3\sqrt{5}}{a}$

$\displaystyle \Rightarrow e^2 = \frac{45}{a^2} \text{ ... ... ... ... ... iii)}$

$\displaystyle \text{Now, } b^2 = a^2 ( e^2 - 1)$

$\displaystyle \Rightarrow 4a = a^2e^2 - a^2$

$\displaystyle \Rightarrow 4a = a^2 \times \frac{45}{a^2} - a^2$

$\displaystyle \Rightarrow a^2 + 4a - 45 = 0$

$\displaystyle \Rightarrow a(a+9) - 5(a+9) = 0$

$\displaystyle \Rightarrow (a-5)(a+9) = 0$

$\displaystyle \Rightarrow a = 5$

$\displaystyle \Rightarrow a^2 = 25$

$\displaystyle \Rightarrow b^2 = 4 \times 5 = 20$

Substituting $\displaystyle a^2 = 25$ and $\displaystyle b^2 = 20$ in equation i) we get the equation of the required hyperbola

$\displaystyle \frac{x^2}{25} - \frac{y^2}{20} = 1$

This is the equation of the required hyperbola.

$\\$

(vii) Given foci $\displaystyle (\pm 4, 0),$ the latus-rectum $\displaystyle = 12$

Since the vertices lie on x-axis, so let the equation of the required hyperbola be

$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \text{ ... ... ... ... ... i)}$

The length of the latus-rectum of the required hyperbola is $\displaystyle 12.$

$\displaystyle \therefore \frac{2b^2}{a} = 12$

$\displaystyle \Rightarrow b^2 = 6a \text{ ... ... ... ... ... ii)}$

Now, the coordinates of foci of the required hyperbola is $\displaystyle ( \pm ae , 0)$

$\displaystyle \therefore ae = 4$

$\displaystyle \Rightarrow e = \frac{4}{a}$

$\displaystyle \Rightarrow e^2 = \frac{16}{a^2} \text{ ... ... ... ... ... iii)}$

$\displaystyle \text{Now, } b^2 = a^2 ( e^2 - 1)$

$\displaystyle \Rightarrow 6a = a^2e^2 - a^2$

$\displaystyle \Rightarrow 6a = a^2 \times \frac{16}{a^2} - a^2$

$\displaystyle \Rightarrow a^2 + 6a - 16 = 0$

$\displaystyle \Rightarrow a(a+8) - 2(a+8) = 0$

$\displaystyle \Rightarrow (a+8)(a-2) = 0 \ \ \ \text{ length cannot be negative } \therefore a+8 \neq 0$

$\displaystyle \Rightarrow a = 2$

$\displaystyle \Rightarrow a^2 = 4$

$\displaystyle \Rightarrow b^2 = 6 \times 2 = 12$

Substituting $\displaystyle a^2 = 4$ and $\displaystyle b^2 = 12$ in equation i) we get the equation of the required hyperbola

$\displaystyle \frac{x^2}{4} - \frac{y^2}{12} = 1$

This is the equation of the required hyperbola.

$\\$

(viii) Given vertices $\displaystyle (0, \pm 6) , e = \frac{5}{3}$

Since the vertices lie on x-axis, so let the equation of the required hyperbola be

$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \text{ ... ... ... ... ... i)}$

The length of the vertices of the required hyperbola is $\displaystyle (\pm a, 0).$

$\displaystyle \therefore a = 7$

$\displaystyle \Rightarrow a^2 = 49 \text{ ... ... ... ... ... ii)}$

$\displaystyle \text{Now, } b^2 = a^2 ( e^2 - 1)$

$\displaystyle \Rightarrow b^2 = 49 \Big[ \Big(\frac{4}{3} \Big)^2 - 1 \Big] \hspace{1.0cm} \Big[ \because e = \frac{4}{3} \Big]$

$\displaystyle \Rightarrow b^2 = 49 \Big[ \frac{16}{9} -1 \Big]$

$\displaystyle \Rightarrow b^2 = 49 \Big[ \frac{7}{9} \Big]$

$\displaystyle \Rightarrow b^2 = \frac{343}{9}$

Substituting $\displaystyle a^2 = 49$ and $\displaystyle b^2 = \frac{343}{9}$ in equation i) we get the equation of the required hyperbola

$\displaystyle \frac{x^2}{49} - \frac{9y^2}{343} = 1$

This is the equation of the required hyperbola.

$\\$

(ix) Given foci $\displaystyle ( 0, \pm \sqrt{10}),$ passing through $\displaystyle (2, 3)$

Since the vertices lie on y-axis, so let the equation of the required hyperbola be

$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = -1 \text{ ... ... ... ... ... i)}$

It passes through $\displaystyle ( 2, 3).$ Therefore

$\displaystyle \frac{2^2}{a^2} - \frac{3^2}{b^2} = -1$

$\displaystyle \Rightarrow \frac{4}{a^2} - \frac{9}{b^2} = -1$

$\displaystyle \Rightarrow \frac{4}{a^2} - \frac{9}{a^2 ( e^2 - 1)} = -1 \text{ ... ... ... ... ... ii)}$

The coordinates of the foci for the required hyperbola are $\displaystyle ( 0, \pm ae)$

$\displaystyle \therefore ae = \sqrt{10}$

$\displaystyle \Rightarrow a^2 e^2 = 10 \text{ ... ... ... ... ... iii)}$

Substituting $\displaystyle a^2 e^2 = 10$ in equation ii) we get

$\displaystyle \frac{4}{a^2} - \frac{9}{10-a^2} = -1$

$\displaystyle \Rightarrow \frac{4(10-a^2) - 9(a^2)}{a^2(10-a^2)} = -1$

$\displaystyle \Rightarrow \frac{40-4a^2-9a^2}{10a^2 - a^4}= -1$

$\displaystyle \Rightarrow 40-13a^2= - 10a^2 + a^4$

$\displaystyle \Rightarrow a^4 + 3a^2 - 40 = 0$

$\displaystyle \Rightarrow a^4 + 8a^2 - 5a^2 - 40 = 0$

$\displaystyle \Rightarrow a^2 ( a^2 + 8) - 5 ( a^2 + 8) = 0$

$\displaystyle \Rightarrow (a^2+8)(a^2-5) = 0$

$\displaystyle \Rightarrow a^2 = 5 \hspace{2.0cm} [ \because a^2 \text{ cannot be negative } ]$

$\displaystyle \text{Now, } b^2 = a^2 ( e^2 -1) = a^2e^2 - a^2 = 10-5 = 5$

Substituting $\displaystyle a^2 = 5$ and $\displaystyle b^2 = 5$ in equation i) we get the equation of the required hyperbola

$\displaystyle \frac{x^2}{5} - \frac{9y^2}{5} = -1$

This is the equation of the required hyperbola.

$\\$

(x) Given foci $\displaystyle ( 0, \pm \sqrt{12}),$ latus-rectum $\displaystyle = 36$

Since the vertices lie on y-axis, so let the equation of the required hyperbola be

$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = -1 \text{ ... ... ... ... ... i)}$

The length of the latus-rectum of the required hyperbola is $\displaystyle 36.$

$\displaystyle \therefore \frac{2a^2}{b} = 36$

$\displaystyle \Rightarrow a^2 = 18b \text{ ... ... ... ... ... ii)}$

The coordinates of foci of the required hyperbola are $\displaystyle ( 0, \pm be)$

$\displaystyle \therefore be = 12$

$\displaystyle \Rightarrow e = \frac{12}{b}$

$\displaystyle \Rightarrow e^2 = \frac{144}{b^2}$

$\displaystyle \text{Now, } a^2 = b^2 ( e^2 - 1)$

$\displaystyle \Rightarrow 18b = b^2 (\frac{144}{b^2} - 1)$

$\displaystyle \Rightarrow 18b = 144 - b^2$

$\displaystyle \Rightarrow b^2 + 18b - 144 = 0$

$\displaystyle \Rightarrow (b-6)(b+24) = 0$

$\displaystyle \Rightarrow b = 6, =24$

Considering positive value of $\displaystyle b = 6$

Therefore $\displaystyle a^2 = 18 ( 6) = 108$

Substituting $\displaystyle a^2 = 108$ and $\displaystyle b^2 = 36$ in equation i) we get the equation of the required hyperbola

$\displaystyle \frac{x^2}{108} - \frac{9y^2}{36} = -1$

$\displaystyle \Rightarrow 3y^2 - x^2 = 108$

This is the equation of the required hyperbola.

$\\$

Question 12: If the distance between the foci of a hyperbola is $\displaystyle 16$ and its eccentricity is $\displaystyle \sqrt{2}$ , then obtain its equation.

Answer:

Eccentricity is $\displaystyle \sqrt{2}$

Distance between foci

$\displaystyle 2ae = 16$

$\displaystyle \Rightarrow 2a (\sqrt{2}) = 16$

$\displaystyle \Rightarrow a = 4\sqrt{2}$

$\displaystyle e = \frac{\sqrt{a^2+b^2}}{a}$

$\displaystyle \sqrt{2} = \frac{\sqrt{32+b^2}}{4\sqrt{2}}$

$\displaystyle 8 = \sqrt{32+b^2}$

$\displaystyle \Rightarrow b^2 = 32$

Therefore the equation of hyperbola is

$\displaystyle \frac{x^2}{32}-\frac{y^2}{32}= 1$

$\displaystyle \Rightarrow x^2 - y^2 = 32$

$\\$

Question 13: Show that the set of all points such that the difference of their distances from $\displaystyle (4,0)$ and $\displaystyle (- 4,0)$ is always equal to $\displaystyle 2$ represents a hyperbola.