Class 11: Parabola – Exercise 25.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the equation of the parabola whose:

(i) focus is $(3, 0)$ and the directrix is $3 x + 4y =1$

(ii) focus is $(1,1)$ and the directrix is $x + y + l =0$

(iii) focus is $(0,0)$ and the directrix $2 x -y -1 = 0$

(iv) focus is $(2,3)$ and the directrix $x -4y + 3=0$

Answer:

(i) Given focus is $(3,0)$ and the directrix is $3 x + 4y =1$

Let $P(x, y)$ be any point on the parabola whose focus is $( 3, 0)$ and the directrix is $3 x + 4y =1$ .

Draw $PM$ perpendicular from $P(x, y)$ on the directrix $3 x + 4y =1$ . Then by definition,

$SP = PM$

$\Rightarrow SP^2 = PM^2$

$\displaystyle \Rightarrow ( x - 3)^2 + ( y-0)^2 = \Bigg| \frac{3 x + 4y -1}{\sqrt{3^2+4^2}} \Bigg|^2$

$\displaystyle \Rightarrow x^2 + 9 - 6x + y^2 = \Bigg| \frac{3 x + 4y -1}{\sqrt{25}} \Bigg|^2$

$\displaystyle \Rightarrow x^2 + 9 - 6x + y^2 = \frac{(3 x + 4y -1)^2}{25}$

$\Rightarrow 25x^2 + 175 - 150 x + 25 y^2 = 9x^2 + 16y^2 +1+24xy -8y-6x$

$\Rightarrow 16x^2 +9y^2 -24xy - 144x + 8y + 224= 0$

This is the equation of the required parabola.

(ii) Given focus is $(1,1)$ and the directrix is $x + y + l =0$

Let $P(x, y)$ be any point on the parabola whose focus is $(1,1)$ and the directrix is $x + y + l =0$ .

Draw $PM$ perpendicular from $P(x, y)$ on the directrix $x + y + l =0$ . Then by definition,

$SP = PM$

$\Rightarrow SP^2 = PM^2$

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

$\displaystyle \Rightarrow x^2 + 1 - 2x + y^2 + 1 - 2y = \Bigg| \frac{x + y +1}{\sqrt{2}} \Bigg|^2$

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

$\Rightarrow 2x^2 + 2 - 4x + 2y^2 + 2 - 4y = x^2 + y^2 +1 + 2xy+ 2x + 2y$

$\Rightarrow x^2 +y^2 -2xy - 6x -6y + 3= 0$

This is the equation of the required parabola.

(iii) Given focus is $(0,0)$ and the directrix is $2 x -y -1=0$

Let $P(x, y)$ be any point on the parabola whose focus is $(0,0)$ and the directrix is $2 x -y -1 = 0$ .

Draw $PM$ perpendicular from $P(x, y)$ on the directrix $2 x -y -1 = 0$ . Then by definition,

$SP = PM$

$\Rightarrow SP^2 = PM^2$

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

$\displaystyle \Rightarrow x^2 + y^2 = \Bigg| \frac{2 x -y -1}{\sqrt{5}} \Bigg|^2$

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

$\Rightarrow 5x^2+5y^2 = 4x^2 + y^2 +1 -4xy- 4x + 2y$

$\Rightarrow x^2 +4y^2 +4xy +4x -2y -1= 0$

This is the equation of the required parabola.

(iv) Given focus is $(2,3)$ and the directrix is $x -4y + 3=0$

Let $P(x, y)$ be any point on the parabola whose focus is $(2,3)$ and the directrix is $x -4y + 3=0$ .

Draw $PM$ perpendicular from $P(x, y)$ on the directrix $x -4y + 3 =0$ . Then by definition,

$SP = PM$

$\Rightarrow SP^2 = PM^2$

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

$\displaystyle \Rightarrow x^2 + 4 - 4x + y^2 + 9 - 6y = \Bigg| \frac{x -4y + 3}{\sqrt{17}} \Bigg|^2$

$\displaystyle \Rightarrow x^2 + 4 - 4x + y^2 + 9 - 6y = \frac{(x -4y + 3)^2}{17}$

$\Rightarrow 17x^2 + 68 - 68x + 17y^2 + 153 - 102y = x^2 +16 y^2 +9 -8xy+ 6x -8y$

$\Rightarrow 16x^2 + y^2 +8xy - 74x -78y + 212= 0$

This is the equation of the required parabola.

$\\$

Question 2: Find the equation of the parabola whose focus is the point $(2, 3)$ and directrix is the line $x - 4 y + 3 = 0$ . Also, find the length of its latus-rectum.

Answer:

Given focus is $(2,3)$ and the directrix is $x -4y + 3=0$

Let $P(x, y)$ be any point on the parabola whose focus is $(2,3)$ and the directrix is $x -4y + 3=0$ .

Draw $PM$ perpendicular from $P(x, y)$ on the directrix $x -4y + 3 =0$ . Then by definition,

$SP = PM$

$\Rightarrow SP^2 = PM^2$

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

$\displaystyle \Rightarrow x^2 + 4 - 4x + y^2 + 9 - 6y = \Bigg| \frac{x -4y + 3}{\sqrt{17}} \Bigg|^2$

$\displaystyle \Rightarrow x^2 + 4 - 4x + y^2 + 9 - 6y =$ $\frac{(x -4y + 3)^2}{17}$

$\Rightarrow 17x^2 + 68 - 68x + 17y^2 + 153 - 102y = x^2 +16 y^2 +9 -8xy+ 6x -8y$

$\Rightarrow 16x^2 + y^2 +8xy - 74x -78y + 212= 0$

This is the equation of the required parabola.

Latus Rectum = length of the perpendicular from ( 2, 3) on $x - 4 y + 3 = 0$

$\displaystyle =2 \Big| \frac{(1)(2) +(-4)(3) + 3}{\sqrt{1^2+(-4)^2}} \Big|$

$\displaystyle =2 \Big| \frac{-7}{\sqrt{17}} \Big|$

$\displaystyle = \frac{14}{\sqrt{17}}$

$\\$

Question 3: Find the equation of the parabola, if

(i) the focus is at $(- 6,- 6)$ and the vertex is at $(- 2,2)$

(ii) the focus is at $(0, - 3)$ and the vertex is at $(0, 0)$

(iii) the focus is at $(0, - 3)$ and the vertex is at $(- 1, - 3)$

(iv) the focus is at $(a, 0)$ arid the vertex is at $(a' , 0)$

(v) the focus is at $(0,0)$ and vertex is at the intersection of the lines $x + y=1$ and $x-y=3$

Answer:

(i) Given focus is at $S(- 6,- 6)$ and the vertex is at $A(- 2,2)$

In order to find the equation of a parabola, we need to know the coordinates of its focus and the equation of the directrix. We are given the coordinates of focus and vertex. So, we require the equation of the directrix. Let $Z(x_1, y_1)$ be the point of intersection of axis and the directrix. The vertex $A$ is the midpoint of the line segment joining the focus $A$ and the point $Z$ of intersection of the axis and directrix. Therefore $A( -2, 2)$ is the midpoint of the line segment joining $S( -6, -6)$ and $Z(x_1, y_1)$ .

Therefore,

$\displaystyle \frac{x_1-6}{2} = 2 \hspace{0.5cm} \Rightarrow x_1 = 2$ and

$\displaystyle \frac{y_1-6}{2} = -2 \Rightarrow y_1 = 10$

Thus the directrix meets the axis at $Z( 2, 10)$

Let $m_1$ be the slope of $AS$ . Then,

$\displaystyle m_1 = \frac{2-(-6)}{-2-(-6)} = \frac{8}{4} = 2$

Let $m_2$ be the slope of the directrix. Since directrix is perpendicular to $AS$ .

$\displaystyle \therefore m_1 m_2 = - 1 \Rightarrow m_2 = \frac{-1}{m_1} = \frac{-1}{2}$

Thus, the directrix passes through the point $Z( 2, 10)$ and has a slope of $\displaystyle \frac{-1}{2}$ . Therefore the equation of the directrix is

$\displaystyle y - 10 = \frac{-1}{2} (x-2)$

$\Rightarrow 2y - 20 = -x + 2$

$\Rightarrow x+ 2y - 22 = 0$

Let $P(x, y)$ be any point on the parabola whose focus is $(- 6,- 6)$ and the directrix is $x+ 2y - 22=0$ .

Draw $PM$ perpendicular from $P(x, y)$ on the directrix $x+ 2y - 22 =0$ . Then by definition,

$SP = PM$

$\Rightarrow SP^2 = PM^2$

$\displaystyle \Rightarrow ( x - (-6))^2 + ( y-(-6))^2 = \Bigg| \frac{ x+ 2y - 22}{\sqrt{1^2+2^2}} \Bigg|^2$

$\displaystyle \Rightarrow x^2 + 36 + 12x + y^2 + 36 +12y = \Bigg| \frac{x+ 2y - 22}{\sqrt{5}} \Bigg|^2$

$\displaystyle \Rightarrow x^2 + 36 + 12x + y^2 + 36 +12y = \frac{(x+ 2y - 22)^2}{5}$

$\Rightarrow 5x^2 + 180 +60x + 5y^2 + 180 +60y = x^2 +4 y^2 +484 +4xy-44x -88y$

$\Rightarrow 4x^2 + y^2 -4xy +104x +148y -124= 0$

This is the equation of the required parabola.

(ii) Given focus is at $S(0, - 3)$ and the vertex is at $A(0, 0)$

In order to find the equation of a parabola, we need to know the coordinates of its focus and the equation of the directrix. We are given the coordinates of focus and vertex. So, we require the equation of the directrix. Let $Z(x_1, y_1)$ be the point of intersection of axis and the directrix. The vertex $A$ is the midpoint of the line segment joining the focus $A$ and the point $Z$ of intersection of the axis and directrix. Therefore $A( 0, 0)$ is the midpoint of the line segment joining $S(0, - 3)$ and $Z(x_1, y_1)$ .

$\displaystyle \text{Therefore } \frac{x_1+0}{2} = 0 \Rightarrow x_1 = 0 \text{ and } \frac{y_1-3}{2} = 0 \Rightarrow y_1 = 3$

Thus the directrix meets the axis at $Z( 0, 3)$

Therefore the equation of directrix is $y = 3$

Let $P(x, y)$ be any point on the parabola whose focus is $(0, -3)$ and the directrix is $y-3=0$ .

Draw $PM$ perpendicular from $P(x, y)$ on the directrix $y-3 =0$ . Then by definition,

$SP = PM$

$\Rightarrow SP^2 = PM^2$

$\displaystyle \Rightarrow ( x - 0)^2 + ( y-(-3))^2 = \Bigg| \frac{ y-3}{\sqrt{0^2+1^2}} \Bigg|^2$

$\displaystyle \Rightarrow x^2 +y^2+ 9 + 6y = \Bigg| \frac{y-3}{1} \Bigg|^2$

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

$\Rightarrow x^2 +12y = 0$

This is the equation of the required parabola.

(iii) Given focus is at $S(0, - 3)$ and the vertex is at $A(- 1, - 3)$

In order to find the equation of a parabola, we need to know the coordinates of its focus and the equation of the directrix. We are given the coordinates of focus and vertex. So, we require the equation of the directrix. Let $Z(x_1, y_1)$ be the point of intersection of axis and the directrix. The vertex $A$ is the midpoint of the line segment joining the focus $A$ and the point $Z$ of intersection of the axis and directrix. Therefore $A(- 1, - 3)$ is the midpoint of the line segment joining $S( 0, - 3)$ and $Z(x_1, y_1)$ .

$\displaystyle \text{Therefore } \frac{x_1+0}{2} = -1 \Rightarrow x_1 = -2 \text{ and } \frac{y_1-3}{2} = -3 \Rightarrow y_1 = -3$

Thus the directrix meets the axis at $Z( -2, -3)$

Let $m_1$ be the slope of $AS$ . Then,

$\displaystyle m_1 = \frac{-3-(-3)}{-1-0} = 0$

Let $m_2$ be the slope of the directrix. Since directrix is perpendicular to $AS$ .

$\displaystyle \therefore m_1 m_2 = - 1 \Rightarrow m_2 = \frac{-1}{m_1} = \frac{-1}{0} = \infty$

Thus, the directrix passes through the point $Z( -2, -3)$ and has a slope of $\infty$ . Therefore the equation of the directrix is

$y - (-3) = \infty (x-(-2))$

$\Rightarrow x+2=0$

Let $P(x, y)$ be any point on the parabola whose focus is $(0, - 3)$ and the directrix is $x+2 = 0$ .

Draw $PM$ perpendicular from $P(x, y)$ on the directrix $x+2 =0$ . Then by definition,

$SP = PM$

$\Rightarrow SP^2 = PM^2$

$\displaystyle \Rightarrow ( x - 0)^2 + ( y-(-3))^2 = \Bigg| \frac{ x+2}{\sqrt{1^2+0^2}} \Bigg|^2$

$\displaystyle \Rightarrow x^2 + y^2 + 9 +6y = \Bigg| \frac{x+ 2}{\sqrt{1}}\Bigg|^2$

$\displaystyle \Rightarrow x^2 + y^2 + 9 +6y = \frac{(x+ 2)^2}{1}$

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

$\Rightarrow y^2-4x+6y+5= 0$

This is the equation of the required parabola.

(iv) Given focus is at $S(a, 0)$ arid the vertex is at $A(a' , 0)$

In order to find the equation of a parabola, we need to know the coordinates of its focus and the equation of the directrix. We are given the coordinates of focus and vertex. So, we require the equation of the directrix. Let $Z(x_1, y_1)$ be the point of intersection of axis and the directrix. The vertex $A$ is the midpoint of the line segment joining the focus $A$ and the point $Z$ of intersection of the axis and directrix. Therefore $A(a' , 0)$ is the midpoint of the line segment joining $S(a, 0)$ and $Z(x_1, y_1)$ .

$\displaystyle \text{Therefore } \frac{x_1+a}{2} = a' \Rightarrow x_1 = 2a'-a \text{ and } \frac{y_1+0}{2} = 0 \Rightarrow y_1 = 0$

Thus the directrix meets the axis at $Z( 2a'-a , 0)$

Let $m_1$ be the slope of $AS$ . Then,

$\displaystyle m_1 = \frac{0-0}{a'-a} = 0$

Let $m_2$ be the slope of the directrix. Since directrix is perpendicular to $AS$ .

$\displaystyle \therefore m_1 m_2 = - 1 \Rightarrow m_2 = \frac{-1}{m_1} = \frac{-1}{0} = \infty$

Thus, the directrix passes through the point $Z( 2a'-a , 0)$ and has a slope of $\infty$ . Therefore the equation of the directrix is

$y - 0 = \infty (x-(2a'-a))$

$\Rightarrow x- (2a'-a) = 0$

Let $P(x, y)$ be any point on the parabola whose focus is $(a, 0)$ and the directrix is $x- (2a'-a) = 0$ .

Draw $PM$ perpendicular from $P(x, y)$ on the directrix $x- (2a'-a) = 0$ . Then by definition,

$SP = PM$

$\Rightarrow SP^2 = PM^2$

$\displaystyle \Rightarrow ( x - a)^2 + ( y-0)^2 = \Bigg| \frac{ x- (2a'-a)}{\sqrt{1^2+0^2}} \Bigg|^2$

$\displaystyle \Rightarrow x^2 +a^2 - 2ax + y^2 = \Bigg| \frac{x- (2a'-a)}{\sqrt{1}} \Bigg|^2$

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

$\Rightarrow x^2 +a^2 - 2ax + y^2 = x^2 + ( 2a'-a)^2 - 2(2a'-a)x$

$\Rightarrow y^2-2ax = 4{a'}^2 - 4a'a - 4a'x + 2ax$

$\Rightarrow y^2 - 4ax + 4a'x - 4{a'}^2 + 4a'a = 0$

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

This is the equation of the required parabola.

(v) Given focus is at $S(0,0)$ and vertex is at the intersection of the lines $x + y=1$ and $x-y=3$

Solving the two equations simultaneously, we vet vertex $A ( 2, -1)$

Given focus is at $S(0,0)$ and the vertex is at $A ( 2, -1)$

In order to find the equation of a parabola, we need to know the coordinates of its focus and the equation of the directrix. We are given the coordinates of focus and vertex. So, we require the equation of the directrix. Let $Z(x_1, y_1)$ be the point of intersection of axis and the directrix. The vertex $A$ is the midpoint of the line segment joining the focus $A$ and the point $Z$ of intersection of the axis and directrix. Therefore $A ( 2, -1)$ is the midpoint of the line segment joining $S(0,0)$ and $Z(x_1, y_1)$ .

$\displaystyle \text{Therefore } \frac{x_1+0}{2} = 2 \Rightarrow x_1 = 4 \text{ and } \frac{y_1+0}{2} = -1 \Rightarrow y_1 = -2$

Thus the directrix meets the axis at $Z( 4, -2)$

Let $m_1$ be the slope of $AS$ . Then,

$\displaystyle m_1 = \frac{-1-0}{2-0} =\frac{-1}{2} = 2$

Let $m_2$ be the slope of the directrix. Since directrix is perpendicular to $AS$ .

$\displaystyle \therefore m_1 m_2 = - 1 \Rightarrow m_2 = \frac{-1}{m_1} =\frac{-1}{\frac{-1}{2}} = 2$

Thus, the directrix passes through the point $Z( 4, -2)$ and has a slope of $2$ . Therefore the equation of the directrix is

$y - (-2) = 2 (x-4)$

$\Rightarrow 2x-y -10=0$

Let $P(x, y)$ be any point on the parabola whose focus is $(0,0)$ and the directrix is $2x-y -10=0$ .

Draw $PM$ perpendicular from $P(x, y)$ on the directrix $2x-y -10=0$ . Then by definition,

$SP = PM$

$\Rightarrow SP^2 = PM^2$

$\displaystyle \Rightarrow ( x - 0)^2 + ( y-0)^2 = \Bigg| \frac{ 2x-y -10}{\sqrt{2^2+(-1)^2}} \Bigg|^2$

$\displaystyle \Rightarrow x^2 + y^2 = \Bigg| \frac{2x-y -10}{\sqrt{5}} \Bigg|^2$

$\displaystyle \Rightarrow x^2 + y^2 = \frac{(2x-y -10)^2}{5}$

$\Rightarrow 5x^2 + 5y^2 = 4x^2 + y^2 + 100 - 4xy - 40x + 20 y$

$\Rightarrow x^2 + 4y^2 +4xy +40x -20y -100= 0$

This is the equation of the required parabola.

$\\$

Question 4: Find the vertex, focus, axis, directrix and latus-rectum of the following parabolas:

i) $y^2 = 8x$

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

iii) $y^2-4y-3x+1=0$

iv) $y^2 - 4y + 4x = 0$

v) $y^2 + 4x + 4y - 3 = 0$

vi) $y^2 = 8x + 8 y$

vii) $4 ( y - 1)^2 = - 7 ( x-3)$

viii) $y^2 = 5x - 4y - 9$

ix) $x^2 + y = 6x - 14$

Note:

	$\displaystyle y^2 = 4ax$	$\displaystyle y^2 = 4ax$	$\displaystyle x^2 = 4ay$	$\displaystyle x^2 = - 4ay$
Coordinates of Vertex	$\displaystyle (0,0)$	$\displaystyle (0,0)$	$\displaystyle (0,0)$	$\displaystyle (0,0)$
Coordinates of focus	$\displaystyle (a, 0)$	$\displaystyle (a, 0)$	$\displaystyle (0, a)$	$\displaystyle (0, -a)$
Equation of the directrix	$\displaystyle x = - a$	$\displaystyle x=a$	$\displaystyle y=-a$	$\displaystyle y=a$
Equation of the axis	$\displaystyle y=0$	$\displaystyle y=0$	$\displaystyle x=0$	$\displaystyle x=0$
Length of Latus rectum	$\displaystyle 4a$	$\displaystyle 4a$	$\displaystyle 4a$	$\displaystyle 4a$
Focal distance of a point $\displaystyle P(x, y)$	$\displaystyle a+x$	$\displaystyle a-x$	$\displaystyle a+y$	$\displaystyle a-y$

Answer:

i) $y^2 = 8x$

The given parabola equation $y^2 = 8x$ is of the form $y^2 = 4ax$ , where $4a = 8 \Rightarrow a = 2$

Therefore:

Vertex: The coordinate of vertex are $(0, 0)$

Focus: The coordinate of vertex are $(a, 0) \text{which is } (2, 0)$

Axes: The equation of the axes is $y = 0$

Directrix: The equation of the axes is $x = -2$

Latus-rectum: The length of the latus-rectum $= 4a = 4 \times 2 = 8 \text{ units}$

$\\$

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

$\displaystyle \text{Given equation } 4x^2+y=0 \text{ is of the form } x^2 = -\frac{1}{4} y.$

$\displaystyle \text{Comparing it with } x^2 = - 4ay \text{ we get } 4a = \frac{1}{4} \Rightarrow a = \frac{1}{16}$

Therefore:

Vertex: The coordinate of vertex are $(0, 0)$

$\displaystyle \text{Focus: The coordinate of vertex are} (0, -a) \text{which is } \Big( 0, \frac{-1}{16} \Big)$

Axes: The equation of the axes is $x = 0$

$\displaystyle \text{Directrix: The equation of the axes is } y = a \Rightarrow y = \frac{1}{16}$

$\displaystyle \text{Latus-rectum: The length of the latus-rectum } = 4a = 4 \times \frac{1}{16} = \frac{1}{4} \text{ units}$

$\\$

iii) The given equation is $\displaystyle y^2-4y-3x+1=0$

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

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

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

Shifting the origin to the point $\displaystyle ( -1, 2)$ without rotating the axes and denoting the new coordinates with respect to $\displaystyle X \text{ and } Y$ , we have

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

Using these relations equation (i), reduces to

$\displaystyle Y^2 = 3X \text{ ... ... ... ... ... (iii)}$

This is of the form $\displaystyle Y^2 = 4aX$

$\displaystyle \text{On comparing, we get } 4a = 3 \Rightarrow a = \frac{3}{4}$

Therefore,

Vertex: The coordinates of the vertex with respect to the new axes are $\displaystyle (X=0, Y=0).$ So the coordinate of the vertex with respect to the old axes are $\displaystyle ( -1, 2).$

$\displaystyle \text{Focus: The coordinates of the focus with respect to the new axes are } \\ \\ (X=\frac{3}{4}, Y=0)$

$\displaystyle \text{Putting } X=\frac{3}{4} \text{ and } Y=0 \text{ in equation (ii), we get}$

$\displaystyle x = \frac{3}{4} -1 = \frac{-1}{4} \text{ and } y = 0+2 = 2$

$\displaystyle \text{Therefore the coordinates of the focus with respect to the old axes are } \Big( \frac{-1}{4}, 2 \Big)$

$\displaystyle \text{Directrix: Equation of directrix is } X = - a \Rightarrow x+1 = \frac{-3}{4} \Rightarrow x = \frac{-7}{4}$

$\displaystyle \text{Axis: Equation of axis is } Y = 0 \Rightarrow y -2 = 0 \Rightarrow y = 2$

$\displaystyle \text{Latus-rectum: Length of latus-rectum } = 4 a = 4 \times \frac{3}{4} = 3 \text{ units}$

$\\$

iv) The given equation is $y^2 - 4y + 4x = 0$

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

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

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

Shifting the origin to the point $\displaystyle ( 1, 2)$ without rotating the axes and denoting the new coordinates with respect to $\displaystyle X \text{ and } Y$ , we have

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

Using these relations equation (i), reduces to

$\displaystyle Y^2 = -4X \text{ ... ... ... ... ... (iii)}$

This is of the form $\displaystyle Y^2 = -4aX$

$\displaystyle \text{On comparing, we get } 4a = 4 \Rightarrow a = 1$

Therefore,

Vertex: The coordinates of the vertex with respect to the new axes are $\displaystyle (X=0, Y=0).$ So the coordinate of the vertex with respect to the old axes are $\displaystyle ( 1, 2).$

$\displaystyle \text{Focus: The coordinates of the focus with respect to the new axes are } \\ \\ (X=-a, Y=0)$

$\displaystyle \text{Putting } x-1=-1 \text{ and } y-2=0 \text{ in equation (ii), we get}$

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

$\displaystyle \text{Therefore the coordinates of the focus with respect to the old axes are } (0, 2)$

$\displaystyle \text{Directrix: Equation of directrix is } X = a \Rightarrow x-1 = 1 \Rightarrow x = 2$

$\displaystyle \text{Axis: Equation of axis is } Y = 0 \Rightarrow y -2 = 0 \Rightarrow y = 2$

$\displaystyle \text{Latus-rectum: Length of latus-rectum } = 4 a = 4 \times 1 = 4 \text{ units}$

$\\$

v) The given equation is $y^2 + 4x + 4y - 3 = 0$

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

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

$\displaystyle \Rightarrow (y+2)^2 = -4x+7$

$\displaystyle \Rightarrow (y+2)^2 = -4 \Big( x-\frac{7}{4} \Big) \text{ ... ... ... ... ... (i)}$

Shifting the origin to the point $\displaystyle ( \frac{7}{4}, -2)$ without rotating the axes and denoting the new coordinates with respect to $\displaystyle X \text{ and } Y$ , we have

$\displaystyle X = x-\frac{7}{4} \text{ and } Y = y + 2 \text{ ... ... ... ... ... (ii)}$

Using these relations equation (i), reduces to

$\displaystyle Y^2 = -4X \text{ ... ... ... ... ... (iii)}$

This is of the form $\displaystyle Y^2 = -4aX$

$\displaystyle \text{On comparing, we get } 4a = 4 \Rightarrow a = 1$

Therefore,

Vertex: The coordinates of the vertex with respect to the new axes are $\displaystyle (X=0, Y=0).$ So the coordinate of the vertex with respect to the old axes are $\displaystyle ( \frac{7}{4}, -2).$

$\displaystyle \text{Focus: The coordinates of the focus with respect to the new axes are } \\ \\ (X=-a = -1, Y=0)$

$\displaystyle \text{Putting } -1 = x-\frac{7}{4} \text{ and } 0 = y+2 \text{ in equation (ii), we get}$

$\displaystyle x = \frac{3}{4} \text{ and } y = -2$

$\displaystyle \text{Therefore the coordinates of the focus with respect to the old axes are } \Big( \frac{3}{4}, -2 \Big)$

$\displaystyle \text{Directrix: Equation of directrix is } X = a \Rightarrow x-\frac{7}{4} = 1 \Rightarrow x = \frac{11}{4}$

$\displaystyle \text{Axis: Equation of axis is } Y = 0 \Rightarrow y +2 = 0 \Rightarrow y = -2$

$\displaystyle \text{Latus-rectum: Length of latus-rectum } = 4 a = 4 \times 1 = 4 \text{ units}$

$\\$

vi) The given equation is $y^2 = 8x + 8 y$

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

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

$\displaystyle \Rightarrow (y-4)^2 = 8(x+2) \text{ ... ... ... ... ... (i)}$

Shifting the origin to the point $\displaystyle (-2, 4)$ without rotating the axes and denoting the new coordinates with respect to $\displaystyle X \text{ and } Y$ , we have

$\displaystyle X = x + 2 \text{ and } Y = y -4 \text{ ... ... ... ... ... (ii)}$

Using these relations equation (i), reduces to

$\displaystyle Y^2 = 8X \text{ ... ... ... ... ... (iii)}$

This is of the form $\displaystyle Y^2 = 4aX$

$\displaystyle \text{On comparing, we get } 4a = 8 \Rightarrow a = 2$

Therefore,

Vertex: The coordinates of the vertex with respect to the new axes are $\displaystyle (X=0, Y=0).$ So the coordinate of the vertex with respect to the old axes are $\displaystyle ( -2, 4).$

$\displaystyle \text{Focus: The coordinates of the focus with respect to the new axes are } \\ \\ (X=a=2, Y=0)$

$\displaystyle \text{Putting } 2 = x+2 \text{ and } 0 = y-4 \text{ in equation (ii), we get}$

$\displaystyle x = 0 \text{ and } y = 4$

$\displaystyle \text{Therefore the coordinates of the focus with respect to the old axes are } (0, 4)$

$\displaystyle \text{Directrix: Equation of directrix is } X = -a \Rightarrow x+2 = -2 \Rightarrow x +4=0$

$\displaystyle \text{Axis: Equation of axis is } Y = 0 \Rightarrow y -4 = 0 \Rightarrow y = 4$

$\displaystyle \text{Latus-rectum: Length of latus-rectum } = 4 a = 4 \times 2 = 8 \text{ units}$

$\\$

vii) The given equation is $4 ( y - 1)^2 = - 7 ( x-3)$

$\displaystyle \Rightarrow ( y - 1)^2 = \frac{-7}{4} ( x-3) \text{ ... ... ... ... ... (i)}$

Shifting the origin to the point $\displaystyle (3,1)$ without rotating the axes and denoting the new coordinates with respect to $\displaystyle X \text{ and } Y$ , we have

$\displaystyle X = x -3 \text{ and } Y = y -1 \text{ ... ... ... ... ... (ii)}$

Using these relations equation (i), reduces to

$\displaystyle Y^2 = \frac{-7}{4}X \text{ ... ... ... ... ... (iii)}$

This is of the form $\displaystyle Y^2 = -4aX$

$\displaystyle \text{On comparing, we get } 4a = \frac{7}{4} \Rightarrow a = \frac{7}{16}$

Therefore,

Vertex: The coordinates of the vertex with respect to the new axes are $\displaystyle (X=0, Y=0).$ So the coordinate of the vertex with respect to the old axes are $\displaystyle ( 3, 1).$

$\displaystyle \text{Focus: The coordinates of the focus with respect to the new axes are } \\ \\ (X=-a, Y=0)$

$\displaystyle \text{Putting }x-3 = \frac{-7}{16} \text{ and } y-1 = 0 \text{ in equation (ii), we get}$

$\displaystyle x =\frac{41}{16} \text{ and } y = 1$

$\displaystyle \text{Therefore the coordinates of the focus with respect to the old axes are } (\frac{41}{16}, 1)$

$\displaystyle \text{Directrix: Equation of directrix is } X = a \Rightarrow x-3 = \frac{7}{16} \Rightarrow x = \frac{55}{16}$

$\displaystyle \text{Axis: Equation of axis is } Y = 0 \Rightarrow y -1 = 0 \Rightarrow y =1$

$\displaystyle \text{Latus-rectum: Length of latus-rectum } = 4 a = 4 \times \frac{7}{16} = \frac{7}{4} \text{ units}$

$\\$

viii) The given equation is $y^2 = 5x - 4y - 9$

$\displaystyle \Rightarrow y^2 + 4y =5x -9$

$\displaystyle \Rightarrow y^2 + 4y+4 =5x -9+4$

$\displaystyle \Rightarrow (y+2)^2 = 5(x-1) \text{ ... ... ... ... ... (i)}$

Shifting the origin to the point $\displaystyle ( 1, -2)$ without rotating the axes and denoting the new coordinates with respect to $\displaystyle X \text{ and } Y$ , we have

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

Using these relations equation (i), reduces to

$\displaystyle Y^2 = 5X \text{ ... ... ... ... ... (iii)}$

This is of the form $\displaystyle Y^2 = 4aX$

$\displaystyle \text{On comparing, we get } 4a = 5 \Rightarrow a = \frac{5}{4}$

Therefore,

Vertex: The coordinates of the vertex with respect to the new axes are $\displaystyle (X=0, Y=0).$ So the coordinate of the vertex with respect to the old axes are $\displaystyle ( 1, -2).$

$\displaystyle \text{Focus: The coordinates of the focus with respect to the new axes are } \\ \\ (X=a, Y=0)$

$\displaystyle \text{Putting } x-1= \frac{5}{4} \text{ and } y+2 = 0 \text{ in equation (ii), we get}$

$\displaystyle x = \frac{9}{4} \text{ and } y = -2$

$\displaystyle \text{Therefore the coordinates of the focus with respect to the old axes are } \Big( \frac{9}{4}, -2 \Big)$

$\displaystyle \text{Directrix: Equation of directrix is } X = -a \Rightarrow x-1 = \frac{-5}{4} \Rightarrow x = \frac{-1}{4}$

$\displaystyle \text{Axis: Equation of axis is } Y = 0 \Rightarrow y +2 = 0 \Rightarrow y = -2$

$\displaystyle \text{Latus-rectum: Length of latus-rectum } = 4 a = 4 \times \frac{5}{4} = 5 \text{ units}$

$\\$

ix) The given equation is $x^2 + y = 6x - 14$

$\displaystyle \Rightarrow x^2-6x= -y-14$

$\displaystyle \Rightarrow x^2-6x+9=-y-14+9$

$\displaystyle \Rightarrow (x-3)^2 = -(y+5) \text{ ... ... ... ... ... (i)}$

Shifting the origin to the point $\displaystyle ( 3, -5)$ without rotating the axes and denoting the new coordinates with respect to $\displaystyle X \text{ and } Y$ , we have

$\displaystyle X = x-3 \text{ and } Y = y + 5 \text{ ... ... ... ... ... (ii)}$

Using these relations equation (i), reduces to

$\displaystyle X^2 = -Y \text{ ... ... ... ... ... (iii)}$

This is of the form $\displaystyle X^2 = -4aY$

$\displaystyle \text{On comparing, we get } 4a = 1 \Rightarrow a = \frac{1}{4}$

Therefore,

Vertex: The coordinates of the vertex with respect to the new axes are $\displaystyle (X=0, Y=0).$ So the coordinate of the vertex with respect to the old axes are $\displaystyle ( 3, -5).$

$\displaystyle \text{Focus: The coordinates of the focus with respect to the new axes are } \\ \\ (X=0, Y=-a)$

$\displaystyle \text{Putting } x-3= 0 \text{ and } y+5 = \frac{1}{4} \text{ in equation (ii), we get}$

$\displaystyle x = 3 \text{ and } y = \frac{-21}{4}$

$\displaystyle \text{Therefore the coordinates of the focus with respect to the old axes are } \Big( 3, \frac{-21}{4} \Big)$

$\displaystyle \text{Directrix: Equation of directrix is } Y = a \Rightarrow y+5 = \frac{1}{4} \Rightarrow y = \frac{-19}{4}$

$\displaystyle \text{Axis: Equation of axis is } X = 0 \Rightarrow x-3 = 0 \Rightarrow x=3$

$\displaystyle \text{Latus-rectum: Length of latus-rectum } = 4 a = 4 \times \frac{1}{4} = 1 \text{ units}$

$\\$

Question 5: For the parabola $y^2 = 4px$ find the extremities of a double ordinate of length $8 p$ . Prove that the lines from the vertex to its extremities are at right angles.

Answer:

Let $\displaystyle PQ$ be the double ordinate of length $\displaystyle 8p$ of the parabola $\displaystyle y^2 = 4px.$

Then, $\displaystyle PR=QR=4p.$

Let $\displaystyle AR = x_1.$ Then, the coordinates of $\displaystyle P \text{ and } Q \text{ are } ( x_1, 4p) \text{ and } (x_1, -4p)$ respectively.

Since $\displaystyle P$ lies on $\displaystyle y^2= 4px.$

$\displaystyle \therefore (4p)^2 = 4px_1 \Rightarrow x_1 = 4p$

So, coordinates of $\displaystyle P \text{ and } Q \text{ are } ( 4p, 4p) \text{ and } ( 4p, -4p)$ respectively.

Also, the coordinates of the vertex $\displaystyle A \text{ are } (0, 0).$

$\displaystyle \therefore m_1 = \text{ Slope of } AP = \frac{4p-0}{4p-0} = 1, \\ \\ \text{ and } , m_2 = \text{ Slope of } AQ = \frac{-4p-0}{4p-0 } = -1$

Clearly, $\displaystyle m_1m_2 = -1.$

Hence $\displaystyle AP$ is perpendicular to $\displaystyle AQ.$

Therefore the lines from the vertex to the extermities are at right angles.

$\\$

Question 6: Find the area of the triangle formed by the lines joining the vertex of the parabola $x^2 =12y$ to the ends of its latus-rectum.

Answer:

The given equation of the parabola: $x^2 =12y$

This is of the form $\displaystyle x^2 = 4ay.$ On comparing we get $\displaystyle 4a = 12 \Rightarrow a = 3$

Therefore the coordinate of the focus $\displaystyle (S)$ is $\displaystyle (0, 3)$

Let $\displaystyle P \text{ and } Q$ lie of the parabola. Therefore,

$\displaystyle x^2 = 12 \times 3 = 36 \Rightarrow x = \pm 6$

Therefore $\displaystyle P ( -6, 3) \text{ and } Q (6, 3).$

$\displaystyle PQ = \sqrt{ (x_2-x_1)^2 + ( y_2-y_1)^2 } = \sqrt{ (6+6)^2+ ( 3-3)^2 } = \sqrt{(12)^2} = 12$

$\displaystyle OS = 3$

Therefore Area of $\displaystyle \triangle POQ = \frac{1}{2} \times 12 \times 3 = 18 \text{sq. units}$

$\\$

Question 7: Find the coordinates of the point of intersection of the axis and the directrix of the parabola whose focus is $( 3, 3)$ and directrix is $3x-4y=2$ . Find also the length of latus-rectum.

Answer:

The equation of directrix is $\displaystyle 3x-4y=2 \text{ ... ... ... ... ... ii)}$

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

Slope of directrix $\displaystyle = \frac{3}{4}$

$\displaystyle \text{We know, axis is perpendicular to directrix.} \\ \\ \text{Therefore slope of axis } = \frac{-1}{\frac{3}{4}} = \frac{-4}{3}$

The focus lies on the axis of parabola, therefore

$\displaystyle (y-3) = \frac{-4}{3} (x-3)$

$\displaystyle \Rightarrow 3y-9 = - 4x + 12$

$\displaystyle \Rightarrow 4x + 3y = 21 \text{ ... ... ... ... ... ii)}$

$\displaystyle \text{Solving i) and ii) we get } x = \frac{18}{5} , y = \frac{11}{5}$

$\displaystyle \text{Therefore the point of intersection of axis and directrix is } ( \frac{18}{5} , \frac{11}{5} )$

$\\$

Question 8: At what point of the parabola $x^2 =9y$ is the, abscissa three times that of ordinate?

Answer:

Let the ordinate of the required point be $\displaystyle y.$

Therefore the abscissa is $\displaystyle 3y.$

Therefore the coordinate of the point is $\displaystyle ( 3y, y).$

This point lie on the parabola $\displaystyle x^2 =9y$ , hence, $\displaystyle (3y, y)$ will satisfy the equation.

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

$\displaystyle \Rightarrow 9y^2 - 9y = 0$

$\displaystyle \Rightarrow y(y-1) = 0$

Since $\displaystyle y \neq 0, y = 1$

Therefore the required point is $\displaystyle ( 3, 1) .$

$\\$

Question 9: Find the equation of a parabola with vertex at the origin, the axis along x-axis and passing through $(2,3)$ .

Answer:

Case 1:

$\displaystyle \text{Let the equation of the required parabola be } y^2 = 4ax$

$\displaystyle \text{Since}, (2, 3) \text{ lies on the equation of the parabola, we get } \\ \\ 3^2 = 4a ( 2) \Rightarrow a = \frac{9}{8}$

$\displaystyle \text{Hence the equation of the parabola is } y^2 = 4 \times \frac{9}{8} x \Rightarrow 2y^2 = 9x$

Case 2:

$\displaystyle \text{Let the equation of the required parabola be } y^2 = -4ax$

$\displaystyle \text{Since}, (2, 3) \text{ lies on the equation of the parabola, we get } \\ \\ 3^2 = -4a ( 2) \Rightarrow a = \frac{-9}{8}$

$\displaystyle \text{Hence the equation of the parabola is } y^2 = -4 \times \frac{-9}{8} x \Rightarrow 2y^2 = 9x$

Either case the equation remains the same which is $\displaystyle 2y^2 = 9x$

$\\$

Question 10: Find the equation of the parabola with vertex at origin and the directrix $y = 2$ .

Answer:

Let $\displaystyle (x_1, y_1)$ be the coordinates of the point of intersection of the axis and the directrix.

Therefore $\displaystyle (x_1, y_1) = ( 0, 2)$

We know that the vertex is the midpoint of the line joining $\displaystyle (0, 2)$ and focus $\displaystyle ( x_2, y_2)$

$\displaystyle \text{Therefore } \frac{x_2+0}{2}= 0 \text{ and } \frac{y_2+2}{2}= 0$

$\displaystyle \Rightarrow x_2 = 0 \text{ and } y_2 = -2$

Therefore the coordinates of the focus is $\displaystyle (0, -2)$

By definition of parabola,

$\displaystyle PS = PM$

$\displaystyle \Rightarrow PS^2 = PM^2$

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

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

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

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

Hence the required equation of the parabola is $\displaystyle x^2 + 8y = 0$

$\\$

Question 11: Find the equation of the parabola whose focus is $(5,2)$ and having vertex at $(3,2)$ .

Answer:

In a parabola, vertex is the midpoint of the focus and the point of intersection of the axis and directrix.

Let $\displaystyle (x_1, y_1)$ be the the coordinates of point of intersection of the axis and directrix.

Therefore $\displaystyle ( 3, 2)$ is the midpoint of the line segment joining $\displaystyle (5, 2)$ and $\displaystyle (x_1, y_1).$

$\displaystyle \text{Therefore } \frac{x_1+5}{2} = 3 \text{ and } \frac{y_1+2}{2}= 2$

$\displaystyle \Rightarrow x_1 + 5 = 6 \text{ and } y_1 + 2 = 4$

$\displaystyle \Rightarrow x_1 = 1 \text{ and } y_1 = 2$

Therefore the directrix meets the axis at $\displaystyle ( 1,2).$

Let $\displaystyle A$ be the vertex and $\displaystyle S$ be the focus of the required parabola.

$\displaystyle \text{Then, } m_1 = \text{slope of } AS = \frac{2-2}{5-3} = 0$

Let $\displaystyle m_2$ be the slope of the directrix. Since directrix is perpendicular to the axis,

$\displaystyle m_2 = \infty$

Thus the directrix passes through $\displaystyle (1, 2)$ with a slope of $\displaystyle \infty$

Therefore the equation is:

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

Let $\displaystyle P(x, y)$ be a point on the parabola.

Then $\displaystyle PS =$ distance of P from directrix

$\displaystyle \sqrt{ (x-5)^2 + (y-2)^2 } = \Big| \frac{x-1}{\sqrt{1^2}} \Big|$

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

$\displaystyle x^2 + 25 - 10 x + y^2 + 4 - 4 y = x^2 + 1 - 2x$

$\displaystyle y^2 - 4y -8x + 28 = 0$

Hence the equation of the required parabola is: $\displaystyle y^2 - 4y -8x + 28 = 0$

$\\$

Question 12: The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wire attached to the cable, the longest wire being 30 m and the shortest wire being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.

Answer:

The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive y-axis.

This can be diagrammatically represented as Here, $\displaystyle AB$ and $\displaystyle OC$ are the longest and the shortest wires, respectively, attached to the cable.

$\displaystyle DF$ is the supporting wire attached to the roadway, $\displaystyle 18 \text{ m }$ from the middle.

$\displaystyle \text{Here}, AB = 30 \text{ m}, OC = 6 \text{ m}, \text{ and } BC =\frac{100}{2} = 50 \text{ m}.$

The equation of the parabola is of the form $\displaystyle x^2 = 4ay$ (as it is opening upwards).

$\displaystyle \text{The coordinates of point A are } (50, 30 - 6) = (50, 24).$

Since $\displaystyle A (50, 24)$ is a point on the parabola,

$\displaystyle (50)^2 = 4a(24)$

$\displaystyle \Rightarrow a = \frac{2500}{4 \times 24} = \frac{625}{24}$

$\displaystyle \text{Therefore the equation of the parabola is } x^2= 4 \times \frac{625}{24} \times y \Rightarrow 6x^2 = 625 y$

The x-coordinate of point $\displaystyle D \text{ is } 18$

$\displaystyle 6(18)^2 = 625 y \Rightarrow y = \frac{6 \times 18 \times 18}{625} \Rightarrow y = 3.11$

Therefore $\displaystyle DE = 3.11$

$\displaystyle DF = DE + EF = 3.11 \text{ m } + 6 \text{ m } = 9.11 \text{ m}$

Thus, the length of the supporting wire attached to the roadway $\displaystyle 18 \text{ m}$ from the middle is approximately $\displaystyle 9.11 \text{ m}$ .

Question 13: Find the equations of the lines joining the vertex of the parabola $y^2 =6x$ to the point on it which have abscissa 24.

Answer:

Let $\displaystyle A$ and $\displaystyle B$ be the points on the parabola $\displaystyle y^2 = 6x$ and $\displaystyle OA$ and $\displaystyle OB$ be the lines joining the vertex $\displaystyle O$ to the points $\displaystyle A$ and $\displaystyle B$ and whose abscissa are $\displaystyle 24.$

Now $\displaystyle y^2 = 6 \times 24 = 144 \Rightarrow y = \pm 12$

$\displaystyle \text{Therefore the coordinates of the points } A \text{ and } B \text{ are } (24, 12 ) \text{and } (24, -12) \\ \\ \text{ respectively. }$

Hence the lines are given by

$\displaystyle y - 0 = \pm \frac{12-0}{24-0} (x-0)$

$\displaystyle \Rightarrow \pm 2y = x$

$\\$

Question 14: Find the coordinates of points on the parabola $y^2 = 8x$ whose focal distance is 4.

Answer:

We have $y^2 = 8x$

$\Rightarrow y^2 = 4 \times 2 x$

Comparing it with general equation of the parabola $y^2 = 4ax,$ we get $a = 2$

Let the required point be $(x_1, y_1)$

Given Focal distance $= 4$

$\Rightarrow x_1 + a = 4 \Rightarrow x_1 = 2$

Now, the point will satisfy the equation of parabola

Therefore $(y_1)^2 = 8 ( 2) = 16 \Rightarrow y_1 = \pm 4$

Hence the coordinate of the points are $(2, 4)$ and $( 2, -4).$

$\\$

Question 15: Find the length of the line segment joining the vertex of the parabola $y^2 =4ax$ and a point on the parabola where the line-segment makes an angle $\theta$ to the x-axis

Answer:

Let the coordinates of the point on the parabola be $\displaystyle B (x_1, y_1)$

Let $\displaystyle BO$ be the line segment.

In the right triangle $\displaystyle AOB:$

$\displaystyle \cos \theta = \frac{AO}{OB} \text{ and } \sin \theta = \frac{AB}{OB}$

$\displaystyle \Rightarrow \cos \theta = \frac{x_1}{OB} \text{ and } \sin \theta = \frac{y_1}{OB}$

Therefore $\displaystyle x_1 = OB \cos \theta \text{ and } y_1 = OB \sin \theta$

Now, the curve is passing through the point $\displaystyle (x_1, y_1)$

Therefore $\displaystyle (y_1)^2 = 4a(x_1)$

$\displaystyle \Rightarrow (OB \sin \theta)^2 = 4a(OB \cos \theta)$

$\displaystyle \Rightarrow OB^2 \sin^2 \theta = 4a OB \cos \theta$

$\displaystyle \Rightarrow OB = \frac{4 a \cos \theta}{\sin^2 \theta} = 4a \mathrm{cosec} \theta \cdot \cot \theta$

$\\$

Question 16: If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.

Answer:

As the vertex and focus lie on y-axis, therefore y-axis is the axis of the parabola.

If the directrix meets the axis of the parabola at point $\displaystyle Z,$ then $\displaystyle AZ = AF = 2$

$\displaystyle OZ = OF + AZ + FA = 2+2+2 = 6$

Hence, the equation of the directrix is $\displaystyle y = 6$

$\displaystyle \text{i.e. } y - 6 =0$

Let $\displaystyle P(x, y)$ be any point in the plane of the focus and directrix and let $\displaystyle MP$ be the perpendicular distance from $\displaystyle P$ to the directrix, then $\displaystyle P$ lies on parabola if and only if $\displaystyle FP = MP$