Class 11: Ellipse – Exercise 26.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note: Also plotted the ellipse equations. Students should have a feel of what the equation looks like.

Question 1: Find the equation of the ellipse whose focus is $\displaystyle (1, - 2),$ the directrix $\displaystyle 3 x - 2y + 5 = 0$ and eccentricity equal to $\displaystyle 1 / 2.$

Answer:

Let $\displaystyle P(x, y)$ be any point on the ellipse whose focus is $\displaystyle S(1, -2)$ and eccentricity $\displaystyle = \frac{1}{2}.$

Let $\displaystyle PM$ be perpendicular from $\displaystyle P$ on directrix. Then,

$\displaystyle SP = e PM$

$\displaystyle \Rightarrow SP = \frac{1}{2} PM$

$\displaystyle \Rightarrow SP^2 = \frac{1}{4} PM^2$

$\displaystyle \Rightarrow 4 SP^2 = PM^2$

$\displaystyle \Rightarrow 4 [ (x-1)^2 + (y+2)^2 ] = \Bigg[ \frac{3x-2y+5}{\sqrt{(3)^2+(-2)^2}} \Bigg]^2$

$\displaystyle \Rightarrow 4 ( x^2 + 1 - 2x + y^2 + 4 + 4y) = \frac{(3x-2y+5)^2}{(\sqrt{13})^2}$

$\displaystyle \Rightarrow 4 ( x^2 + y^2 -2x+4y+5) = \frac{(3x-2y+5)^2}{13}$

$\displaystyle \Rightarrow 52( x^2 + y^2 -2x+4y+5) = (3x-2y+5)^2$

$\displaystyle \Rightarrow 52x^2+52y^2-104x+208y+260 = 9x^2+4y^2+25-12xy-20y+30x$

$\displaystyle \Rightarrow 52x^2-9x^2+52y^2-4y^2+12xy-104x-30x+208y+20y+260-25=0$

$\displaystyle \Rightarrow 43x^2+48y^2+12xy-134x+228y+235=0$

This is the required equation of the ellipse.

$\\$

Question 2: Find the equation of the ellipse in the following cases:

$\displaystyle \text{(i) focus is } (0, 1), \text{ directrix is } x + y= 0 \text{ and } e = \frac{1}{2}$

$\displaystyle \text{(ii) focus is } (- 1, 1), \text{ directrix is } x -y+ 3 =0 \text{ and } e = \frac{1}{2}$

$\displaystyle \text{(iii) focus is } (- 2,3), \text{ directrix is } 2 x + 3 y + 4= 0 \text{ and } e= \frac{4}{5}$

$\displaystyle \text{(iv) focus is } (1,2), \text{ directrix is } 3 r + 4y -5= 0 \text{ and } e = \frac{1}{2}$

Answer:

(i) Let $\displaystyle P(x, y)$ be a point on ellipse. Then, by definition

$\displaystyle SP = e PM$

$\displaystyle \text{Given focus is } (0, 1), \text{ directrix is } x + y= 0 \text{ and } e = \frac{1}{2}$

$\displaystyle SP = \frac{1}{2} PM$

$\displaystyle \Rightarrow SP^2 = \frac{1}{4} PM^2$

$\displaystyle \Rightarrow 4SP^2 = PM^2$

$\displaystyle \Rightarrow 4[ (x-0)^2 + ( y-1)^2 ] = \Bigg[ \frac{x+y}{\sqrt{1^2+1^2} } \Bigg]^2$

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

$\displaystyle \Rightarrow 8( x^2 + y^2 + 1 - 2y ) = x^2 + y^2 + 2xy$

$\displaystyle \Rightarrow 8x^2 + 8y^2 + 8 - 16y = x^2 + y^2 + 2xy$

$\displaystyle \Rightarrow 7x^2 + 7y^2 - 2xy - 16y + 8 = 0$

This is the required equation of the ellipse.

(ii) Let $\displaystyle P(x, y)$ be a point on ellipse. Then, by definition

$\displaystyle SP = e PM$

$\displaystyle \text{Given focus is } (- 1, 1), \text{ directrix is } x -y+ 3 =0 \text{ and } e = \frac{1}{2}$

$\displaystyle SP = \frac{1}{2} PM$

$\displaystyle \Rightarrow SP^2 = \frac{1}{4} PM^2$

$\displaystyle \Rightarrow 4SP^2 = PM^2$

$\displaystyle \Rightarrow 4[ (x+1)^2 + ( y-1)^2 ] = \Bigg[ \frac{x-y+3}{\sqrt{(1)^2+(-1)^2} } \Bigg]^2$

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

$\displaystyle \Rightarrow 8( x^2 + y^2 +2x - 2y+2 ) = x^2+y^2+9-6y-2xy+6x$

$\displaystyle \Rightarrow 8x^2-x^2+8y^2-y^2+2xy+16x-6x-16y+6y+16-9=0$

$\displaystyle \Rightarrow 7x^2 + 7y^2 + 2xy +10x- 10y + 7 = 0$

This is the required equation of the ellipse.

(iii) Let $\displaystyle P(x, y)$ be a point on ellipse. Then, by definition

$\displaystyle SP = e PM$

$\displaystyle \text{Given focus is } (- 2,3), \text{ directrix is } 2 x + 3 y + 4= 0 \text{ and } e= \frac{4}{5}$

$\displaystyle SP = \frac{4}{5} PM$

$\displaystyle \Rightarrow SP^2 = \frac{16}{25} PM^2$

$\displaystyle \Rightarrow 25SP^2 = 16PM^2$

$\displaystyle \Rightarrow 25[ (x+2)^2 + ( y-3)^2 ] = 16 \Bigg[ \frac{2x+3y+4}{\sqrt{(2)^2+(3)^2} } \Bigg]^2$

$\displaystyle \Rightarrow 25( x^2 + 4+4x+y^2 + 9 - 6y ) = \frac{16(2x+3y+4)^2}{13}$

$\displaystyle \Rightarrow 325( x^2+4+4x+y^2+9-6y ) = 16( 4x^2+9y^2+16+12xy+24y+16x )$

$\displaystyle \Rightarrow 325x^2+1300+1300x+325y^2+2925-1950y= 64x^2+144y^2+256+192xy+384y+256x$

$\displaystyle \Rightarrow 261x^2+181y^2+1044x-2309y-192xy+3969=0$

This is the required equation of the ellipse.

(iv) Let $\displaystyle P(x, y)$ be a point on ellipse. Then, by definition

$\displaystyle SP = e PM$

$\displaystyle \text{Given focus is } (1,2), \text{ directrix is } 3 r + 4y -5= 0 \text{ and } e = \frac{1}{2}$

$\displaystyle SP = \frac{1}{2} PM$

$\displaystyle \Rightarrow SP^2 = \frac{1}{4} PM^2$

$\displaystyle \Rightarrow 4SP^2 = PM^2$

$\displaystyle \Rightarrow 4[ (x-1)^2 + ( y-2)^2 ] = \Bigg[ \frac{3x+4y-5}{\sqrt{(3)^2+(4)^2} } \Bigg]^2$

$\displaystyle \Rightarrow 4( x^2 + 1-2x+y^2 + 4 - 4y ) = \frac{(3x+4y-5)^2}{25}$

$\displaystyle \Rightarrow 100( x^2 + y^2 -2x - 4y+5 ) = 9x^2+16y^2+25+24xy-40y-30x$

$\displaystyle \Rightarrow 100x^2 + 100y^2 -200x - 400y+500 = 9x^2+16y^2+25+24xy-40y-30x$

$\displaystyle \Rightarrow 91x^2 + 84y^2 - 24xy -170x- 360y + 475 = 0$

This is the required equation of the ellipse.

$\\$

Question 3: Find the eccentricity, coordinates of foci, length of the latus-rectum of the following ellipse:

(i) $\displaystyle 4 x^2 +9 y^2 =1$

(ii) $\displaystyle 5x^2 + 4y^2 =1$

(iii) $\displaystyle 4x^2 + 3y^2 =1$

(iv) $\displaystyle 25 x^2 + 16 y^2 =1600$

(v) $\displaystyle 9x^2 + 25y^2 - 225$

Answer:

(i) $\displaystyle \text{Given } 4 x^2 +9 y^2 =1$

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

$\displaystyle \text{This is of the form } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \\ \\ \text{ where } a^2 = \frac{1}{4} \text{ and } b^2= \frac{1}{9}, \text{ i.e. } a = \frac{1}{2} \text{ and } b = \frac{1}{3}$

$\displaystyle \text{Clearly } a > b$

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

$\displaystyle \Rightarrow e = \sqrt{1 - \frac{\frac{1}{9}}{\frac{1}{4}}}$

$\displaystyle \Rightarrow e = \sqrt{1 - \frac{4}{9}}$

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

$\displaystyle \text{Coordinates of the foci } = ( \pm ae, 0) = \Bigg( \pm \frac{\sqrt{5}}{3}, 0 \Bigg)$

$\displaystyle \text{Length of the latus rectum } = \frac{2b^2}{a} = \frac{2 \times \frac{1}{9}}{\frac{1}{2} } = \frac{4}{9}$

(ii) $\displaystyle \text{Given } 5x^2 + 4y^2 =1$

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

$\displaystyle \text{This is of the form } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \\ \\ \text{ where } a^2 = \frac{1}{5} \text{ and } b^2= \frac{1}{4}, \text{ i.e. } a = \frac{1}{\sqrt{5}} \text{ and } b = \frac{1}{2}$

$\displaystyle \text{Clearly } b > a$

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

$\displaystyle \Rightarrow e = \sqrt{1 - \frac{\frac{1}{5}}{\frac{1}{4}}}$

$\displaystyle \Rightarrow e = \sqrt{1 - \frac{4}{5}}$

$\displaystyle \Rightarrow e = \frac{1}{\sqrt{5}}$

$\displaystyle \text{Coordinates of the foci } = ( 0, \pm be) = \Bigg( 0, \pm \frac{1}{2\sqrt{5}} \Bigg)$

$\displaystyle \text{Length of the latus rectum } = \frac{2a^2}{b} = \frac{2 \times \frac{1}{5}}{\frac{1}{2} } = \frac{4}{5}$

(iii) $\displaystyle \text{Given } 4x^2 + 3y^2 =1$

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

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

$\displaystyle \text{Clearly } b > a$

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

$\displaystyle \Rightarrow e = \sqrt{1 - \frac{\frac{1}{4}}{\frac{1}{3}}}$

$\displaystyle \Rightarrow e = \sqrt{1 - \frac{3}{4}}$

$\displaystyle \Rightarrow e = \frac{1}{2}$

$\displaystyle \text{Coordinates of the foci } = ( 0, \pm be) = \Bigg( 0, \pm \frac{1}{2\sqrt{3}} \Bigg)$

$\displaystyle \text{Length of the latus rectum } = \frac{2a^2}{b} = \frac{2 \times \frac{1}{4}}{\frac{1}{\sqrt{3}} } = \frac{\sqrt{3}}{2}$

(iv) $\displaystyle \text{Given } 25 x^2 + 16 y^2 =1600$

$\displaystyle \Rightarrow \frac{x^2}{64} + \frac{y^2}{100} = 1$

$\displaystyle \text{This is of the form } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \\ \\ \text{ where } a^2 = 64 \text{ and } b^2= 100, \text{ i.e. } a = 8 \text{ and } b = 10$

$\displaystyle \text{Clearly } b > a$

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

$\displaystyle \Rightarrow e = \sqrt{1 - \frac{64}{100}}$

$\displaystyle \Rightarrow e = \sqrt{\frac{36}{100}}$

$\displaystyle \Rightarrow e = \frac{6}{10} \text{ or } \frac{3}{5}$

$\displaystyle \text{Coordinates of the foci } = ( 0, \pm be) = ( 0, \pm 6 )$

$\displaystyle \text{Length of the latus rectum } = \frac{2a^2}{b} = \frac{2 \times 64}{10 } = \frac{64}{5}$

(v) $\displaystyle \text{Given } 9x^2 + 25y^2 - 225$

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

$\displaystyle \text{This is of the form } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \\ \\ \text{ where } a^2 = 25 \text{ and } b^2= 9, \text{ i.e. } a = 5 \text{ and } b = 3$

$\displaystyle \text{Clearly } a > b$

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

$\displaystyle \Rightarrow e = \sqrt{1 - \frac{9}{25}}$

$\displaystyle \Rightarrow e = \sqrt{\frac{16}{25}}$

$\displaystyle \Rightarrow e = \frac{4}{5}$

$\displaystyle \text{Coordinates of the foci } = ( \pm ae, 0) = ( \pm 4, 0 )$

$\displaystyle \text{Length of the latus rectum } = \frac{2b^2}{a} = \frac{2 \times 9}{5 } = \frac{18}{5}$

$\\$

Question 4: Find the equation to the ellipse (referred to its axes as the axes of $\displaystyle x$ and $\displaystyle y$ respectively which passes through the point $\displaystyle (- 3, 1)$ and has eccentricity $\displaystyle \sqrt{\frac{2}{5}}$

Answer:

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

$\displaystyle \text{We know } e = \sqrt{1 - \frac{b^2}{a^2} }$

$\displaystyle \text{Given } e = \sqrt{\frac{2}{5}}$

$\displaystyle \Rightarrow \sqrt{\frac{2}{5}} = \sqrt{1 - \frac{b^2}{a^2}}$

$\displaystyle \Rightarrow \frac{2}{5} = 1 - \frac{b^2}{a^2}$

$\displaystyle \Rightarrow \frac{b^2}{a^2} = 1 - \frac{2}{5}$

$\displaystyle \Rightarrow \frac{b^2}{a^2} = \frac{3}{5}$

$\displaystyle \Rightarrow 5b^2 = 3a^2$

$\displaystyle \Rightarrow b^2 = \frac{3a^2}{5} \text{ ... ... ... ... ... ii) }$

Substituting this in equation i) we get

$\displaystyle \frac{9}{a^2} + \frac{1}{\frac{3a^2}{5} } = 1$

$\displaystyle \Rightarrow \frac{9}{a^2} + \frac{5}{3a^2 } = 1$

$\displaystyle \Rightarrow \frac{1}{a^2} \Big[ 9 + \frac{5}{3} \Big] = 1$

$\displaystyle \Rightarrow \frac{32}{3} = a^2 \text{ ... ... ... ... ... iii) }$

$\displaystyle \text{Substituting } a^2 = \frac{32}{3} \text{in equation ii) we get}$

$\displaystyle b^2 = \frac{3}{5} \times \frac{32}{3} = \frac{32}{5} \text{ ... ... ... ... ... iv) }$

Therefore the equation of ellipse is

$\displaystyle \frac{x^2}{\frac{32}{3}} + \frac{y^2}{\frac{32}{5}} = 1$

$\displaystyle \Rightarrow \frac{3x^2}{32} + \frac{5y^2}{32} = 1$

$\displaystyle \Rightarrow 3x^2 + 5y^2 = 32$

This is the required equation of ellipse.

$\\$

Question 5: Find the equation of the ellipse in the following cases:

$\displaystyle \text{(i) eccentricity } e = \frac{1}{2} \text{ and foci } (\pm 2, 0)$

$\displaystyle \text{(ii) eccentricity } e = \frac{2}{3} \text{ and length of latus-rectum } = 5$

$\displaystyle \text{(iii) eccentricity } e = \frac{1}{2} \text{ and semi-major axis } = 4$

$\displaystyle \text{(iv) eccentricity } e = \frac{1}{2} \text{ and maior axis } = 12$

$\displaystyle \text{(v) The ellipse passes through } (1,4) \text{ and } (- 6,1).$

$\displaystyle \text{(vi) Vertices } (\pm 5,0), \text{ foci } (\pm 4, 0)$

$\displaystyle \text{(vii) Vertices } (0, \pm 13), \text{ foci } (0, \pm 5)$

$\displaystyle \text{(viii) Vertices } (\pm 6, 0), \text{ foci } (\pm 4, 0)$

$\displaystyle \text{(ix) Ends of major axis } (\pm 3,0), \text{ ends of minor axis } (0, \pm 2)$

$\displaystyle \text{(x) Ends of major axis } (0, \pm \sqrt{5}), \text{ ends of minor axis } (\pm1, 0)$

$\displaystyle \text{(xi) Length of major axis } 26, \text{ foci } (\pm 5, 0)$

$\displaystyle \text{(xii) Length of minor axis } 16 \text{ foci } (0, \pm 6)$

$\displaystyle \text{(xiii) Foci } (\pm 3, 0), a = 4$

Answer:

$\displaystyle \text{(i) Given eccentricity } e = \frac{1}{2} \text{ and foci } (\pm 2, 0)$

Coordinates of the foci $\displaystyle = ( \pm ae, 0)$

$\displaystyle \text{We have } ae = 2 \Rightarrow a \times \frac{1}{2} = 2 \Rightarrow a = 4$

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

$\displaystyle \Rightarrow \frac{1}{2} = \sqrt{ 1 - \frac{b^2}{16} }$

Squaring both sides, we get:

$\displaystyle \frac{1}{4} = \frac{16-b^2}{16}$

$\displaystyle \Rightarrow 4 = 16-b^2$

$\displaystyle \Rightarrow b^2 = 12$

$\displaystyle \text{Now, } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\displaystyle \Rightarrow \frac{x^2}{16} + \frac{y^2}{12} = 1$

$\displaystyle \Rightarrow 3x^2 + 4y^2 = 48$

This is the required equation of the ellipse.

$\displaystyle \text{(ii) Given eccentricity } e = \frac{2}{3} \text{ and length of latus-rectum } = 5$

$\displaystyle \text{We have } \frac{2b^2}{a} = 5 \Rightarrow 2b^2 = 5a \Rightarrow b^2 = \frac{5a}{2}$

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

$\displaystyle \Rightarrow \frac{2}{3} = \sqrt{ 1 - \frac{\frac{5a}{2}}{a^2} }$

Squaring both sides, we get:

$\displaystyle \frac{4}{9} = \frac{2a-5}{2a}$

$\displaystyle \Rightarrow 8a = 18a-45$

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

$\displaystyle \therefore b^2 = \frac{45}{4}$

$\displaystyle \text{Now, } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\displaystyle \Rightarrow \frac{4x^2}{81} + \frac{4y^2}{45} = 1$

$\displaystyle \Rightarrow 20x^2 + 36y^2 = 405$

This is the required equation of the ellipse.

$\displaystyle \text{(iii) Given eccentricity } e = \frac{1}{2} \text{ and semi-major axis } = 4$

$\displaystyle i.e \ a = 4$

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

$\displaystyle \Rightarrow \frac{1}{2} = \sqrt{ 1 - \frac{b^2}{16} }$

Squaring both sides, we get:

$\displaystyle \frac{1}{4} = \frac{16-b^2}{16}$

$\displaystyle \Rightarrow 16 = 64-4b^2$

$\displaystyle \Rightarrow b^2 = 12$

$\displaystyle \text{Now, } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\displaystyle \Rightarrow \frac{x^2}{16} + \frac{y^2}{12} = 1$

$\displaystyle \Rightarrow 3x^2 + 4y^2 = 48$

This is the required equation of the ellipse.

$\displaystyle \text{(iv) Given eccentricity } e = \frac{1}{2} \text{ and major axis } = 12$

$\displaystyle i.e \ 2a = 12 \ or \ a = 6$

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

$\displaystyle \Rightarrow \frac{1}{2} = \sqrt{ 1 - \frac{b^2}{36} }$

Squaring both sides, we get:

$\displaystyle \frac{1}{4} = \frac{36-b^2}{36}$

$\displaystyle \Rightarrow 36 = 144-4b^2$

$\displaystyle \Rightarrow b^2 = 27$

$\displaystyle \text{Now, } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\displaystyle \Rightarrow \frac{x^2}{36} + \frac{y^2}{27} = 1$

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

This is the required equation of the ellipse.

$\displaystyle \text{(v) Given The ellipse passes through } (1,4) \text{ and } (- 6,1).$

Ellipse passes through $\displaystyle ( 1, 4)$

$\displaystyle \text{Now, } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

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

$\displaystyle \text{Let } \frac{1}{a^2} = \alpha \text{ and } \frac{1}{b^2} = \beta$

$\displaystyle \text{Then } \alpha + 16 \beta = 1 \text{ ... ... ... ... ... i) }$

$\displaystyle \text{Similarly, Ellipse passes through } ( -6, 1)$

$\displaystyle \text{Now, } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\displaystyle \Rightarrow \frac{36}{a^2} + \frac{1}{b^2} = 1$

$\displaystyle \therefore 36\alpha + \beta = 1 \text{ ... ... ... ... ... ii) }$

Solving i) and ii) we get

$\displaystyle \alpha = \frac{3}{115} \text{ and } \beta = \frac{7}{115}$

Substituting the values, we get:

$\displaystyle \frac{3x^2}{115} + \frac{7y^2}{115} = 1$

$\displaystyle \Rightarrow \frac{3x^2 + 7y^2}{115} = 1$

$\displaystyle \Rightarrow 3x^2 + 7y^2 = 115$

This is the required equation of the ellipse.

$\displaystyle \text{(vi) Given Vertices } (\pm 5,0), \text{ foci } (\pm 4, 0)$

Let the equation of the required ellipse be

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

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

$\displaystyle \therefore a = 5 \text{ and } ae = 4 \Rightarrow e = \frac{4}{5}$

$\displaystyle \text{Now, } b^2 = a^2 ( 1 - e^2) \Rightarrow b^2 = 25 \Bigg( 1 - \frac{16}{25} \Bigg) = 9$

Substituting the values of $\displaystyle a^2$ and $\displaystyle b^2$ in ( i) we get,

$\displaystyle \frac{x^2}{25} + \frac{y^2}{9} = 1$

This is the required equation of the ellipse.

$\displaystyle \text{(vii) Given Vertices } (0, \pm 13), \text{ foci } (0, \pm 5)$

Let the equation of the required ellipse be

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

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

$\displaystyle \therefore b = 13 \text{ and } be = 5 \Rightarrow e = \frac{5}{13}$

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

Substituting the values of $\displaystyle a^2$ and $\displaystyle b^2$ in ( i) we get,

$\displaystyle \frac{x^2}{144} + \frac{y^2}{169} = 1$

This is the required equation of the ellipse.

$\displaystyle \text{(viii) Given Vertices } (\pm 6, 0), \text{ foci } (\pm 4, 0)$

Let the equation of the required ellipse be

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

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

$\displaystyle \therefore a = 6 \text{ and } ae = 4 \Rightarrow e = \frac{2}{3}$

$\displaystyle \text{Now, } b^2 = a^2 ( 1 - e^2) \Rightarrow b^2 = 36 \Bigg( 1 - \frac{16}{36} \Bigg) = 20$

Substituting the values of $\displaystyle a^2$ and $\displaystyle b^2$ in ( i) we get,

$\displaystyle \frac{x^2}{36} + \frac{y^2}{20} = 1$

This is the required equation of the ellipse.

$\displaystyle \text{(ix) Given Ends of major axis } (\pm 3,0), \text{ ends of minor axis } (0, \pm 2)$

Let the equation of the required ellipse be

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

End of major axis $\displaystyle = (\pm 3, 0)$

End of minor axis $\displaystyle = ( 0, \pm 2)$

The coordinates of the end points of the major and minor axes are $\displaystyle ( \pm a, 0)$ and $\displaystyle ( 0, \pm b)$ respectively.

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

$\displaystyle \text{Therefore } \frac{x^2}{9} + \frac{y^2}{4} = 1$

This is the required equation of the ellipse.

$\displaystyle \text{(x) Ends of major axis } (0, \pm \sqrt{5}), \text{ ends of minor axis } (\pm1, 0)$

Let the equation of the required ellipse be

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

End of major axis $\displaystyle = (0, \pm \sqrt{5} )$

End of minor axis $\displaystyle = ( \pm 1, 0)$

The coordinates of the end points of the major and minor axes are $\displaystyle ( \pm a, 0)$ and $\displaystyle ( 0, \pm b)$ respectively.

$\displaystyle \therefore a = 1 \text{ and } b = \sqrt{5}$

$\displaystyle \text{Therefore } \frac{x^2}{1} + \frac{y^2}{5} = 1$

This is the required equation of the ellipse.

$\displaystyle \text{(xi) Given Length of major axis } 26, \text{ foci } (\pm 5, 0)$

Let the equation of the required ellipse be

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\displaystyle \text{We have } 2a = 26 \Rightarrow a = 13$

$\displaystyle \text{Also, } ae = 5 \Rightarrow e = \frac{5}{13}$

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

$\displaystyle \Rightarrow \frac{5}{13} = \sqrt{ 1 - \frac{b^2}{169} }$

Squaring both sides we get:

$\displaystyle \frac{25}{169} = \frac{169-b^2}{169}$

$\displaystyle \Rightarrow b^2 = 144$

$\displaystyle \text{Therefore } \frac{x^2}{169} + \frac{y^2}{144} = 1$

This is the required equation of the ellipse.

$\displaystyle \text{(xii) Given Length of minor axis } 16 \text{ foci } (0, \pm 6)$

Let the equation of the required ellipse be

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\displaystyle \text{We have } 2b = 16 \Rightarrow b = 8$

$\displaystyle \text{Also, } be = 6 \Rightarrow e = \frac{6}{8}$

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

$\displaystyle \Rightarrow \frac{6}{8} = \sqrt{ 1 - \frac{a^2}{64} }$

Squaring both sides we get:

$\displaystyle \frac{36}{64} = \frac{64-a^2}{64}$

$\displaystyle \Rightarrow a^2 = 28$

$\displaystyle \text{Therefore } \frac{x^2}{64} + \frac{y^2}{28} = 1$

This is the required equation of the ellipse.

$\displaystyle \text{(xiii) Given Foci } (\pm 3, 0), a = 4$

$\displaystyle \text{ i.e } ae = 3 \Rightarrow e = \frac{3}{4}$

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

$\displaystyle \Rightarrow \frac{3}{4} = \sqrt{ 1 - \frac{b^2}{16} }$

Squaring both sides we get:

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

$\displaystyle \Rightarrow b^2 = 7$

$\displaystyle \text{Therefore } \frac{x^2}{16} + \frac{y^2}{7} = 1$

This is the required equation of the ellipse.

$\\$

Question 6: Find the equation of the eclipse whose foci are (4,0) and ( -4, 0) , eccentricity $\displaystyle = \frac{1}{3}.$

Answer:

Let the equation of the required ellipse be

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

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

Also $\displaystyle e = \frac{1}{3}$

We have $\displaystyle ae = 4 \Rightarrow a \times \frac{1}{3} = 4 \Rightarrow a = 12$

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

$\displaystyle \Rightarrow b^2 = 12^2 \Bigg(1 - \Bigg( \frac{1}{3} \Bigg) \Bigg)^2$

$\displaystyle \Rightarrow b^2 = 144 ( \frac{8}{9} )$

$\displaystyle \Rightarrow b^2 = 128$

Substituting the values of $\displaystyle a^2$ and $\displaystyle b^2$ in equation (i), we get:

$\displaystyle \frac{x^2}{144} + \frac{y^2}{128} = 1$

This is the required equation of the ellipse.

$\\$

Question 7: Find the equation of the-ellipse in the standard form whose minor axis is equal to the distance between foci and whose latus-rectum is 10.

Answer:

Given minor axis is equal to the distance between foci and whose latus-rectum is 10.

$\displaystyle i.e \frac{2b^2}{a} = 10 \Rightarrow b^2 = 5a$

$\displaystyle \text{ and } 2b = 2ae \Rightarrow b = ae \Rightarrow b^2 = a^2 e^2$

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

$\displaystyle \Rightarrow b^2 = a^2 - b^2$

$\displaystyle \Rightarrow a^2 = 10a ( \because b^2 = 5a)$

$\displaystyle \Rightarrow a = 10$

$\displaystyle \Rightarrow b^2 = 5a$

$\displaystyle \Rightarrow b^2 = 50$

Substituting the values of a and b in the equation of an ellipse, we get:

$\displaystyle \frac{x^2}{100} + \frac{y^2}{50} = 1$

$\displaystyle \Rightarrow x^2 + 2y^2 = 100$

This is the required equation of the ellipse.

$\\$

Question 8: Find the equation of the ellipse whose center is (- 2,3) and whose semi-axis are 3 and 2 when major axis is (i) parallel to x-axis (ii) parallel to y-axis

Answer:

(i) Let $\displaystyle 2a$ and $\displaystyle 2b$ be the major and minor axes of the ellipse. Then, the equation is

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

Given center is $\displaystyle (2, 3)$

We have semi major axis $\displaystyle = a = 3$ $\displaystyle \Rightarrow a^2 = 9$

Similarly, semi major axis $\displaystyle = b = 2$ $\displaystyle \Rightarrow b^2 = 4$

Putting $\displaystyle a^2 = 9$ and $\displaystyle b^2 = 4$ in equation (i) we get

$\displaystyle \frac{(x+2)^2}{9} + \frac{(y-3)^2}{4} = 1$

$\displaystyle \Rightarrow \frac{4(x+2)^2+9(y-3)^2}{36} = 1$

$\displaystyle \Rightarrow 4(x+2)^2 + 9(y-3)^2 = 36$

$\displaystyle \Rightarrow 4(x^2 + 4+ 4x) + 9( y^2+9-6y) = 36$

$\displaystyle \Rightarrow 4x^2 + 16 + 16x + 9y^2+81-54y = 36$

$\displaystyle \Rightarrow 4x^2 + 9y^2 + 16x - 54y + 61 = 0$

This is the required equation of the ellipse.

(ii) Let $\displaystyle 2a$ and $\displaystyle 2b$ be the major and minor axes of the ellipse. Then, the equation is

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

We have semi major axis $\displaystyle = a = 2$ $\displaystyle \Rightarrow a^2 = 4$

Similarly, semi major axis $\displaystyle = b = 3$ $\displaystyle \Rightarrow b^2 = 9$

Given center is $\displaystyle (2, 3)$

Putting $\displaystyle a^2 = 4$ and $\displaystyle b^2 = 9$ in equation (i) we get

$\displaystyle \frac{(x+2)^2}{4} + \frac{(y-3)^2}{9} = 1$

$\displaystyle \Rightarrow \frac{9(x+2)^2+4(y-3)^2}{36} = 1$

$\displaystyle \Rightarrow 9(x+2)^2 + 4(y-3)^2 = 36$

$\displaystyle \Rightarrow 9(x^2 + 4+ 4x) + 4( y^2+9-6y) = 36$

$\displaystyle \Rightarrow 9x^2 + 36 + 36x + 4y^2+36-24y = 36$

$\displaystyle \Rightarrow 9x^2 + 4y^2 + 36x - 24y + 36 = 0$

This is the required equation of the ellipse.

$\\$

Question 9: Find the eccentricity of an ellipse whose latus-rectum is:

(i) half of its minor axis (ii) half of its major axis.

Answer:

Let $\displaystyle 2a$ and $\displaystyle 2b$ be the major and minor axes of the ellipse.

(i) When latus-rectum is half of minor axis

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

$\displaystyle \Rightarrow 2b^2 = ab$

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

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

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

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

$\displaystyle \Rightarrow \frac{1}{4} = 1-e^2$

$\displaystyle \Rightarrow e^2 = 1 - \frac{1}{4}$

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

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

(ii) When latus-rectum is half of minor axis

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

$\displaystyle \Rightarrow 2b^2 = a^2$

$\displaystyle \Rightarrow a^2 = 2b^2$

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

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

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

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

$\displaystyle \Rightarrow e^2 = \frac{1}{2}$

$\displaystyle \Rightarrow e = \frac{1}{\sqrt{2}}$

$\\$

Question 10: Find the center, the lengths of the axes, eccentricity, foci of the following ellipse:

$\displaystyle (i) x^2 + 2y^2 -2x+12y + 10 = 0$

$\displaystyle (ii) x^2 + 4y^2 - 4x + 24 y + 31 = 0$

$\displaystyle (iii) 4x^2 + y^2 - 8x + 2y + 1 = 0$

$\displaystyle (iv) 3x^2 + 4y^2- 12x - 8y + 4 = 0$

$\displaystyle (v) 4x^2 + 16y^2 - 24x - 32 y -12 = 0$

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

Answer:

$\displaystyle \text{(i) Given: } x^2 + 2y^2 -2x+12y + 10 = 0$

$\displaystyle \Rightarrow (x^2-2x)+2(y^2+6y) = - 10$

$\displaystyle \Rightarrow (x^2-2x+1)+2(y^2+6y+9)= -10+18+1$

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

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

Here, $\displaystyle x_1 = 1$ and $\displaystyle y_1 = - 3$

$\displaystyle \text{Also, } a = 3 \text{ and } b = \frac{3}{\sqrt{2}}$

The coordinates of the center of the ellipse are $\displaystyle (x_1, y_1) = ( 1, -3)$

Clearly, $a > b$ so, the given equation represents an ellipse whose major and minor axes are along the X and Y axes respectively.

$\displaystyle \text{Major axis } = 2a \Rightarrow 2 \times 3 = 6$

$\displaystyle \text{Minor axis } = 2b \Rightarrow 2 \times \frac{3}{\sqrt{2}} = 3 \sqrt{2}$

Eccentricity: The eccentricity $e$ is given by

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

$\displaystyle \Rightarrow e = \sqrt{1 - \frac{\frac{9}{2}}{9}}$

$\displaystyle \Rightarrow e = \frac{1}{\sqrt{2}}$

$\displaystyle \text{Foci } = (x_1 \pm ae, y_1) = (1 \pm \frac{3}{\sqrt{2}}, -3)$

$\displaystyle \text{(ii) Given: } x^2 + 4y^2 - 4x + 24 y + 31 = 0$

$\displaystyle \Rightarrow (x^2-4x)+4(y^2+6y) = - 31$

$\displaystyle \Rightarrow (x^2-4x+4)+4(y^2+6y+9)= -31+36+4$

$\displaystyle \Rightarrow (x-2)^2+ 4 (y+3)^2 = 9$

$\displaystyle \Rightarrow \frac{(x-2)^2}{9} + \frac{(y+3)^2}{\frac{9}{4}} = 1$

Here, $\displaystyle x_1 = 2$ and $\displaystyle y_1 = - 3$

$\displaystyle \text{Also, } a = 3 \text{ and } b = \frac{3}{2}$

The coordinates of the center of the ellipse are $\displaystyle (x_1, y_1) = ( 2, -3)$

Clearly, $a > b$ so, the given equation represents an ellipse whose major and minor axes are along the X and Y axes respectively.

$\displaystyle \text{Major axis } = 2a \Rightarrow 2 \times 3 = 6$

$\displaystyle \text{Minor axis } = 2b \Rightarrow 2 \times \frac{3}{2} = 3$

Eccentricity: The eccentricity $e$ is given by

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

$\displaystyle \Rightarrow e = \sqrt{1 - \frac{\frac{9}{4}}{9}}$

$\displaystyle \Rightarrow e = \frac{\sqrt{3}}{2}$

$\displaystyle \text{Foci } = (x_1 \pm ae, y_1) = (2 \pm \frac{3\sqrt{3}}{2}, -3)$

$\displaystyle \text{(iii) Given: } 4x^2 + y^2 - 8x + 2y + 1 = 0$

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

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

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

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

Here, $\displaystyle x_1 = 1$ and $\displaystyle y_1 = - 1$

$\displaystyle \text{Also, } a = 1 \text{ and } b = 2$

The coordinates of the center of the ellipse are $\displaystyle (x_1, y_1) = ( 1, -1)$

Clearly, $b > a$ so, the given equation represents an ellipse whose major and minor axes are along the Y and X axes respectively.

$\displaystyle \text{Major axis } = 2b \Rightarrow 2 \times 2 = 4$

$\displaystyle \text{Minor axis } = 2a \Rightarrow 2 \times 1 = 2$

Eccentricity: The eccentricity $e$ is given by

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

$\displaystyle \Rightarrow e = \sqrt{1 - \frac{1}{4}}$

$\displaystyle \Rightarrow e = \frac{\sqrt{3}}{2}$

$\displaystyle \text{Foci } = (x_1 , y_1\pm be) = (2 , -1 \pm \sqrt{3})$

$\displaystyle \text{(iv) Given: } 3x^2 + 4y^2- 12x - 8y + 4 = 0$

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

$\displaystyle \Rightarrow 3(x^2-4x+4)+4(y^2-2y+1)= -4+12+4$

$\displaystyle \Rightarrow 3(x-2)^2+ 4 (y-1)^2 = 12$

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

Here, $\displaystyle x_1 = 2$ and $\displaystyle y_1 = 1$

$\displaystyle \text{Also, } a = 2 \text{ and } b = \sqrt{3}$

The coordinates of the center of the ellipse are $\displaystyle (x_1, y_1) = ( 2, 1)$

Clearly, $a > b$ so, the given equation represents an ellipse whose major and minor axes are along the X and Y axes respectively.

$\displaystyle \text{Major axis } = 2a \Rightarrow 2 \times 2 = 4$

$\displaystyle \text{Minor axis } = 2b \Rightarrow 2 \times \sqrt{3} = 2\sqrt{3}$

Eccentricity: The eccentricity $e$ is given by

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

$\displaystyle \Rightarrow e = \sqrt{1 - \frac{3}{4}}$

$\displaystyle \Rightarrow e = \frac{1}{2}$

$\displaystyle \text{Foci } = (x_1 \pm ae, y_1) = (2 \pm 1, 1)$

$\displaystyle \text{(v) Given: } 4x^2 + 16y^2 - 24x - 32 y -12 = 0$

$\displaystyle \Rightarrow 4(x^2-6x)+16(y^2-2y) = 12$

$\displaystyle \Rightarrow 4(x^2-6x+9)+16(y^2-2y+1)= 12+36+16$

$\displaystyle \Rightarrow 4(x-3)^2+ 16 (y-1)^2 = 64$

$\displaystyle \Rightarrow \frac{(x-3)^2}{16} + \frac{(y-1)^2}{4} = 9$

Here, $\displaystyle x_1 = 3$ and $\displaystyle y_1 = 1$

$\displaystyle \text{Also, } a = 4 \text{ and } b = 2$

The coordinates of the center of the ellipse are $\displaystyle (x_1, y_1) = ( 3, 1)$

Clearly, $a > b$ so, the given equation represents an ellipse whose major and minor axes are along the X and Y axes respectively.

$\displaystyle \text{Major axis } = 2a \Rightarrow 2 \times 4 = 8$

$\displaystyle \text{Minor axis } = 2b \Rightarrow 2 \times 2 = 4$

Eccentricity: The eccentricity $e$ is given by

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

$\displaystyle \Rightarrow e = \sqrt{1 - \frac{4}{16}}$

$\displaystyle \Rightarrow e = \frac{\sqrt{3}}{2}$

$\displaystyle \text{Foci } = (x_1 \pm ae, y_1) = (3 \pm 2\sqrt{3}, 1)$

$\displaystyle \text{(vi) Given: } x^2 + 4y^2 - 2x = 0$

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

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

$\displaystyle \Rightarrow (x-1)^2+ 4 (y-0)^2 = 1$

$\displaystyle \Rightarrow \frac{(x-1)^2}{1} + \frac{(y-0)^2}{\frac{1}{4}} = 1$

Here, $\displaystyle x_1 = 1$ and $\displaystyle y_1 = 0$

$\displaystyle \text{Also, } a = 1 \text{ and } b = \frac{1}{2}$

The coordinates of the center of the ellipse are $\displaystyle (x_1, y_1) = ( 1, 0)$

Clearly, $a > b$ so, the given equation represents an ellipse whose major and minor axes are along the X and Y axes respectively.

$\displaystyle \text{Major axis } = 2a \Rightarrow 2 \times 1 = 2$

$\displaystyle \text{Minor axis } = 2b \Rightarrow 2 \times \frac{1}{2} = 1$

Eccentricity: The eccentricity $e$ is given by

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

$\displaystyle \Rightarrow e = \sqrt{1 - \frac{\frac{1}{4}}{1}}$

$\displaystyle \Rightarrow e = \frac{\sqrt{3}}{2}$

$\displaystyle \text{Foci } = (x_1 \pm ae, y_1) = (1 \pm\frac{\sqrt{3}}{2}, 1)$

$\\$

Question 11: Find the equation of an ellipse whose foci are at $\displaystyle (\pm 3,0)$ and which passes through $\displaystyle (4, 1).$

Answer:

Let the equation of the required ellipse be

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

The coordinates of its foci are $\displaystyle ( \pm ae, 0) \text{ i.e. } (\pm 3, 0)$

$\displaystyle \therefore ae = 3$

$\displaystyle \Rightarrow (ae)^2 = 9 \text{ ... ... ... ... ... ii) }$

The required ellipse passes through $\displaystyle (4, 1)$

$\displaystyle \text{Therefore } \frac{4^2}{a^2}+ \frac{1^2}{b^2} = 1$

$\displaystyle \Rightarrow 16b^2 + a^2 = a^2b^2$

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

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

$\displaystyle \Rightarrow b^2 = a^2 - 9$

Substituting this in equation ii) we get

$\displaystyle a^2 + 16( a^2 - 9) = a^2 ( a^2 - 9)$

$\displaystyle \Rightarrow a^2 + 16a^2 - 144 = a^4 - 9a^2$

$\displaystyle 17a^2 - 144 = a^4 - 9a^2$

$\displaystyle a^2 - 26a^2 + 144 = 0$

$\displaystyle (a^2 - 18)(a^2 - 8) = 0$

$\displaystyle a^2 = 18 \text{ or } a^2 \neq 8$ (because then $b^2$ would be negative which is not possible).

Therefore $\displaystyle b^2 = 18 - 9 = 9$

Therefore the required equation of the ellipse is

$\displaystyle \frac{x^2}{18}+ \frac{y^2}{9} = 1$

This is the required equation of the ellipse.

$\\$

Question 12: Find the equation of an ellipse whose eccentricity is $\displaystyle 2/3,$ latus-rectum is $\displaystyle 5$ and the center is at the origin.

Answer:

Let the equation of the required ellipse be

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

$\displaystyle \text{The length of latus-rectum is } = 5$

$\displaystyle \therefore \frac{2b^2}{a} = 5$

$\displaystyle \Rightarrow b^2 = \frac{5a}{2} \text{ ... ... ... ... ... ii)}$

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

$\displaystyle \text{Given } e = \frac{2}{3}$

$\displaystyle \Rightarrow \frac{5a}{2} = a^2 \Big[ 1 - \Big( \frac{2}{3} \Big)^2 \Big]$

$\displaystyle \Rightarrow \frac{5}{2} = a \Big(\frac{5}{9} \Big)$

$\displaystyle \Rightarrow \frac{5}{2} \times \frac{9}{5} = a$

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

$\displaystyle \Rightarrow a^2 = \frac{81}{4}$

$\displaystyle \text{Substituting } a = \frac{9}{2} \text{ in } b^2 = \frac{5a}{2}, \text{ we get }$

$\displaystyle b^2 = \frac{5}{2} \times \frac{9}{2}$

$\displaystyle \Rightarrow b^2 = \frac{45}{4}$

$\displaystyle \text{Substituting } a^2 = \frac{81}{4} \text{ and } b^2 = \frac{45}{4} \text{ in equation i) we get }$

$\displaystyle \frac{x^2}{\frac{81}{4}}+ \frac{y^2}{\frac{45}{4}} = 1$

$\displaystyle \frac{4x^2}{81}+ \frac{4y^2}{45} = 1$

This is the required equation of the ellipse.

$\\$

Question 13: Find the equation of an ellipse with its foci on y-axis, eccentricity $\displaystyle 3/4,$ center at the origin and passing through $\displaystyle (6,4).$

Answer:

Let the equation of the ellipse be

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

$\displaystyle \text{Given } e = \frac{3}{4}$

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

$\displaystyle \Rightarrow a^2 = b^2 [ 1 - (\frac{3}{4})^2]$

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

$\displaystyle \Rightarrow a^2 = b^2 ( \frac{7}{16}) \text{ ... ... ... ... ... ii)}$

$\displaystyle \text{The required ellipse passes through } (6, 4)$

$\displaystyle \therefore \frac{6^2}{a^2}+\frac{4^2}{b^2} = 1$

$\displaystyle \Rightarrow \frac{36}{a^2}+\frac{16}{b^2} = 1$

$\displaystyle \Rightarrow \frac{36}{\frac{7}{16} b^2}+\frac{16}{b^2} = 1$

$\displaystyle \Rightarrow \frac{576}{7b^2} + \frac{16}{b^2} = 1$

$\displaystyle \Rightarrow \frac{1}{b^2} [ \frac{576}{7} + \frac{16}{1} ] = 1$

$\displaystyle \Rightarrow b^2 = \frac{688}{7}$

$\displaystyle \text{Substituting } b^2 = \frac{688}{7} \text{ in equation ii) we get }$

$\displaystyle a^2 = \frac{7}{16} \times \frac{688}{7}$

$\displaystyle \Rightarrow a^2 = 43$

Substituting in equation i) we get

$\displaystyle \frac{x^2}{43}+\frac{y^2}{\frac{688}{7}} = 1$

$\displaystyle \Rightarrow \frac{x^2}{43}+\frac{7y^2}{688} = 1$

This is the equation of the required ellipse.

$\\$

Question 14: Find the equation of an ellipse whose axes lie along coordinate axes and which passes through $\displaystyle (4, 3)$ and $\displaystyle (-1, 4).$

Answer:

Let the equation of the required ellipse be

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

The required ellipse passes through $\displaystyle (4, 3)$ and $\displaystyle (-1, 4)$

$\displaystyle \therefore \frac{4^2}{a^2}+ \frac{3^2}{b^2} = 1$

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

$\displaystyle \Rightarrow 16b^2 + 9a^2 = a^2b^2 \text{ ... ... ... ... ... ii)}$

$\displaystyle \text{Similarly, } \frac{(-1)^2}{a^2}+ \frac{4^2}{b^2} = 1$

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

$\displaystyle \Rightarrow 16a^2 + b^2 = a^2b^2 \text{ ... ... ... ... ... iii)}$

Solving equation ii) and iii) we get

$\displaystyle a^2 = \frac{247}{7} \text{ and } b^2 = \frac{247}{15}$

Substituting in equation i) we get

$\displaystyle \frac{x^2}{\frac{247}{7}}+ \frac{y^2}{\frac{247}{15}} = 1$

$\displaystyle \Rightarrow \frac{7x^2}{247}+ \frac{15y^2}{247} = 1$

This is the equation of the required ellipse.

$\\$

Question 15: Find the equation of an ellipse whose axes lie along the coordinate axes, which passes through the point $\displaystyle (-3, 1)$ and has eccentricity equal to $\displaystyle \sqrt{2/5}.$

Answer:

Let the equation of the required ellipse be

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

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

$\displaystyle \Rightarrow b^2 = a^2 \Bigg[ 1 - \Bigg( \sqrt{\frac{2}{5}} \Bigg)^2 \Bigg]$

$\displaystyle \Rightarrow b^2 = a^2 \Bigg[ 1 - \frac{2}{5} \Bigg]$

$\displaystyle \Rightarrow b^2 = a^2 \times \frac{3}{5}$

$\displaystyle \Rightarrow b^2 = \frac{3a^2}{5} \text{ ... ... ... ... ... ii)}$

The required ellipse passes through $\displaystyle ( -3, -1)$

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

$\displaystyle \Rightarrow \frac{9}{a^2}+ \frac{1}{b^2} = 1 \text{ ... ... ... ... ... iii)}$

Substituting ii) into iii) we get

$\displaystyle \frac{9}{a^2}+ \frac{1}{\frac{3a^2}{5}} = 1$

$\displaystyle \Rightarrow \frac{1}{a^2} \Bigg[\frac{9}{1}+\frac{5}{3} \Bigg] = 1$

$\displaystyle \Rightarrow a^2 = \frac{32}{3}$

Substituting in ii) we get

$\displaystyle b^2 = \frac{3}{5} \times \frac{32}{3} = \frac{32}{5}$

Substituting in i) we get

$\displaystyle \frac{x^2}{\frac{32}{3}}+ \frac{y^2}{\frac{32}{5}} = 1$

$\displaystyle \Rightarrow \frac{3x^2}{32}+ \frac{5y^2}{32} = 1$

$\displaystyle \Rightarrow 3x^2 + 5y^2 = 32$

This is the equation of the required ellipse.

$\\$

Question 16: Find the equation of an ellipse, the distance between the foci is 8 units and the distance between the directrices is 18 units.

Answer:

Let the equation of the required ellipse be

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

The distance between foci is $\displaystyle 8$ units

$\displaystyle \text{i.e. } 2ae = 8 \text{ ... ... ... ... ... ii)}$

The distance between the directrics is $\displaystyle 18$ units

$\displaystyle \text{i.e. } \frac{2a}{e} = 18 \text{ ... ... ... ... ... iii)}$

From ii) and iii) we get $\displaystyle a^2 = 36$

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

$\displaystyle \Rightarrow b^2 = 36 - (ae)^2 = 36 - 16 = 20$

Substituting $\displaystyle a^2 = 36$ and $\displaystyle b^2 = 20$ in equation i) we get

$\displaystyle \frac{x^2}{36}+ \frac{y^2}{20} = 1$

This is the equation of the required ellipse.

$\\$

Question 17: Find the equation of an ellipse whose vertices are $\displaystyle (0, \pm 10)$ and eccentricity $\displaystyle e = \frac{4}{5}$

Answer:

Let the equation of the required ellipse be

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

The coordinates of the vertices are $\displaystyle (0, \pm b) \text{ i.e } (1, \pm 10)$

$\displaystyle \therefore b = 10$

$\displaystyle \Rightarrow b^2 = 100$

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

$\displaystyle \Rightarrow a^2 = 100 \Big[ 1 - \Big(\frac{4}{5} \Big)^2 \Big]$

$\displaystyle \Rightarrow a^2 = 100 \Big[ 1- \frac{16}{25} \Big]$

$\displaystyle \Rightarrow a^2 = 100 \Big[ \frac{9}{25} \Big] = 36$

Substituting $\displaystyle a^2 = 36$ and $\displaystyle b^2 = 100$ in equation i) we get

$\displaystyle \frac{x^2}{36}+ \frac{y^2}{100} = 1$

$\displaystyle \Rightarrow 100x^2 + 36y^2 = 3600$

This is the equation of the required ellipse.

$\\$

Question 18: A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with x-axis.

Answer:

Let $\displaystyle AB$ be the rod making an angle $\displaystyle \theta$ with $\displaystyle OX$ and let $\displaystyle P$ be the point on it such that $\displaystyle AP = 3 \text{ cm. }$

$\displaystyle AB = 12 \text{ cm }$

$\displaystyle \text{Then } PB = AB - AP = (12-3) = 9 \text{ cm }$

$\displaystyle \text{From } P, \text{ draw } PQ \perp OY \text{ and } PR \perp OX$

$\displaystyle \text{In } \triangle PBQ, \text{ we have }$

$\displaystyle \cos \theta = \frac{PQ}{PB} = \frac{x}{9}$

$\displaystyle \text{In } \triangle PRA, \text{we have }$

$\displaystyle \sin \theta = \frac{PR}{PA} = \frac{y}{3}$

$\displaystyle \text{Since } \sin^2 \theta + \cos^2 \theta = 1, \text{ we have: }$

$\displaystyle \Big( \frac{y}{3} \Big)^2 + \Big( \frac{x}{9 } \Big)^2 = 1$

$\displaystyle \Rightarrow \frac{x^2}{81} + \frac{y^2}{9} = 1$

$\displaystyle \text{Thus the locus of a point } P \text{ on the rod is } \frac{x^2}{81} + \frac{y^2}{9} = 1$

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Question 19: Find the equation of the set of all points whose distances from $\displaystyle (0,4)$ are $\displaystyle 2/3$ of their distances from the line $\displaystyle y = 9.$