Class 12 Parabola Exercise 3.2.1 Solutions | Pioneer Mathematics Dreamland Publication Pvt. Ltd.

The given values of the vertex (h,k) = (0,0) and focus (a,0) = (-3/2,0) of the parabola suggest that it is a parabola whose axis is the x-axis.

So, its equation is y^2 = 4ax.

By comparing the values of x-coordinates of the focus, we get, $\rm a = - \frac{3}[2}$.

Hence, the required equation of the parabola is:

$\rm y^2 = 4 \cdot \left ( - \frac{3}{2} \right ) x$

$\rm or, y^2 = - 2 \cdot 3 x$

$\rm \therefore y^2 = -6x$

The focal length of the parabola is $\rm 4a = \pm \left [ 4 \left ( - \frac{3}{2} \right ) \right ]$

$\rm = \pm \left [ 2 \cdot (-3) \right ]$

$\rm = \pm \left [ -6 \right ]$

$\rm = 6$ units

The parabola opens left as $\rm a = - \frac{3}{2} < 0$.

The given parabola has a vertex (h,k) = (0,0). So, its equation is either $\rm y^2 = 4ax$ or $\rm x^2 = 4ay$.

Here, the coordinates of the focus of the parabola are given (0,4).

The x-coordinates of both the vertex and the parabola are the same. So, the axis of the parabola is the Y-axis. The coordinates of the focus of such a parabola are represented by S (0, a) = (0,4).

Here, $\rm a = 4$

Now, the equation of a parabola is

$\rm x^2 = 4ay$

$\rm or, x^2 = 4 \cdot 4 \cdot y$

$\rm \therefore x^2 = 16y$

And, the focal length of the parabola is $\rm l = 4a = 4 \cdot 4 = 16$ units.

The parabola opens upward because the value of a > 0.

The vertex of the parabola has a x-coordinate of 5. Similarly, the x-coordinate of the focus of the parabola is 5. It suggests that the axis of the parabola is parallel to the y-axis. So, its equation is $\rm (x - h)^2 = 4a(y-k)$.

We know, the coordinates of the focus is S (h, k + a). Given, vertex (h,k) = (5,3). And, S (h, k+a) = (5,6). 

On comparing, $\rm k + a = 6$

$\rm or, 3 + a = 6$

$\rm or, a = 6 - 3$

$\rm \therefore a = 3$

Now, put the values of h,k, and a in the suggested equation to get the equation of the parabola.

$\rm (x - 5)^2 = 4 \cdot 3 \cdot (y - 3)$

$\rm \therefore (x - 5)^2 = 12 ( y - 3)$

Hence, the required equation of the parabola has been obtained.

Its focal length = a = 3 units.

The parabola opens upwards as a = 3, a > 0.

Let (h,k) be the vertex of the parabola. Let (x,y) be the passing point on the parabola. Let a be the focal distance of the parabola.

We have,

(h,k) = (0,0)

(x,y) = (5,2)

The parabola is symmetric about the y-axis. This means that the y-axis is the axis of the parabola. Our desired equation of a parabola whose axis is y-axis is $\rm (x - h)^2 = 4a(y - k)$.

Put the values of (h,k) and (x,y) in the above equation to find the value of a, we get,

$\rm (5 - 0)^2 = 4a (2 - 0)$

$\rm or, 5^2 = 4a \cdot 2$

$\rm or, 25 = 8a$

$\rm \therefore a = \frac{25}{8}$

Hence, the required equation of the parabola is given by

$\rm (x - h)^2 = 4a (y - k)$

$\rm or, (x - 0)^2 = 4 \cdot \frac{25}{8} (y - 0)$

$\rm or, x^2 = \frac{25}{2} ( y)$

$\rm \therefore 2 x^2 = 25y$

The required equation of the parabola is obtained. It is $\rm 2x^2 = 25y$.

The length of the latus rectum of the parabola is 16. $\rm 4a = 16$

So, we get $\rm a = 4$

The axis of the parabola is parallel to the x-axis. This means that the equation of the parabola is $\rm (y - k)^2 = 4a(x - h)$, where (h,k) is the coordinates of its vertex.

The parabola passes through points (3,2) and (3,-2). Its axis lies on the x-axis. The average of the y-coordinates of the parabola when it passes through the same x-coordinates gives the y-coordinate of the vertex of the parabola.

$\rm k = \frac{2 + (-2)}{2} = \frac{0}{2}$

$\rm \therefore k = 0$

Put the value of 4a = 16 and k = 0 in the equation of the parabola, and we get,

$\rm ( y - 0)^2 = 16 (x - h)$ – (1)

We need to find the value of h.

As (3,2) is a passing point of the parabola, we can replace the values of (x,y) = (3,2) in the above equation and solve for h, we get,

$\rm (2 - 0)^2 = 16 (3 - h)$

$\rm or, 2^2 = 48 - 16h$

$\rm or, 4 = 48 - 16h$

$\rm or, 16h = 48 - 4$

$\rm or, 16h = 44$

$\rm or, h = \frac{44}{16}$

$\rm \therefore h = \frac{11}{4}$

Now, replace the value of h in the equation (1) of the parabola, we get,

$\rm ( y  - 0)^2 = 16 \left ( x - \frac{11}{4} \right )$

$\rm or, y^2 = 16 \left ( \frac{4x - 11}{4} \right )$

$\rm \therefore y^2 = 4 \cdot (4x - 11)$

Hence, the required equation of the parabola is $\rm y^2 = 4 (4x - 11)$.

The given parabola $\rm x^2 - 4x - 3y + 13 = 0$ follows the following equation of parabola:

$$\rm (x - h)^2 = 4a (y-k)$$

where (h,k) represents the vertex of the parabola.

To find the coordinates of the vertex, we factorize the given equation of the parabola

$\rm x^2 - 4x - 3y + 13 = 0$

$\rm or, x^2 - 4x + 13 = 3y$

$\rm or, x^2 - 4x + (4 - 4) + 13 = 3y$

$\rm or, x^2 - 4x + 4 + 9 = 3y$

$\rm or, x^2 - 2 \cdot 2 \cdot x + 2^2 = 3y - 9$

$\rm or, (x - 2)^2 = 3 (y - 3)$

$\rm or, (x - 2)^2 = \frac{4}{4} \cdot 3 \cdot (y - 3)$

$\rm \therefore (x - 2)^2 = 4 \cdot \frac{3}{4} ( y - 3)$

Now, we compare this equation with the above equation to get the values of (h,k). So, $\rm (h, k) = (2,3)$.

Also, $\rm a = \frac{3}{4}$.

The coordinates of the focus are given by (h, k + a). So, the coordinates of the focus is

$\rm (h, k+a) = \left ( 2, 3 + \frac{3}{4} \right )$

$\rm = \left ( 2 , \frac{12 + 3}{4} \right )$

$\rm = \left (2, \frac{15}{4} \right )$

The equation of the axis of the parabola is given by

$\rm x= h$

$\rm \therefore x = 2$

The equation of the directrix of the parabola is given by

$\rm y = k - a$

$\rm or, y = 3 - \frac{3}{4}$

$\rm or, y = \frac{12 - 3}{4}$

$\rm or, y = \frac{9}{4}$

$\rm or, 4y = 9$

$\rm \therefore 4y - 9 = 0$ is the required equation of the directrix of the parabola.

The length of the latus rectum is given by $\rm 4a = 4 \cdot \frac{3}{4} = 3$ units.

We know, that the latus rectum passes through the focus of the parabola. For the given parabola, the y-coordinates of the focus are the same as the y-coordinates of the ends of the latus rectum.

Let us assume that the point P $\rm (x_1, k+a)$ and $\rm (x_2, k+a)$ are the coordinates of the ends of the latus rectum. We know, $\rm k + a = \frac{15}{4}$.

Put the value of $\rm y = k + a$ in the equation of the parabola to get the two values of x, we get,

$\rm (x - 2)^2 =3 (y - 3)$

$\rm or, (x - 2)^2 = 3 \left ( \frac{15}{4} - 3 \right )$

$\rm or, (x - 2)^2 = 3 \cdot \left ( \frac{15 - 12}{4} \right )$

$\rm or, (x - 2)^2 = 3 \cdot \frac{3}{4}$

$\rm or, (x - 2)^2 = \frac{9}{4}$

$\rm or, (x - 2) = \pm \sqrt{ \frac{9}{4} }$

$\rm or, x - 2 = \pm \frac{3}{2}$

Taking positive sign, we get,

$\rm or, x - 2 = \frac{3}{2}$

$\rm or, x = \frac{3}{2} + 2$

$\rm or, x = \frac{3 + 4}{2}$

$\rm \therefore x= \frac{7}{2}$

Let this value of x be the value of $\rm x_1$. So, $\rm x_1 = \frac{7}{2}$.

Taking negative sign, we get,

$\rm or, x - 2 = - \frac{3}{2}$

$\rm or, x = - \frac{3}{2} + 2$

$\rm or, x = - \frac{3 - 4}{2}$

$\rm or, x = \frac{1}{2}$

$\rm \therefore x = \frac{1}{2}$

Let this value of x be the value of $\rm x_2$. So, $\rm x_2 = \frac{1}{2}$.

Hence, the required coordinates of the ends of the latus rectum are $\rm \left ( \frac{7}{2}, \frac{15}{4} \right )$ and $\rm \left ( \frac{1}{2}, \frac{15}{4} \right )$

The focal distance of point P(2, 4) is the distance from P to the focus of the parabola S(a, 0).

The given equation of the parabola is $\rm y^2 = 8x$.

$\rm or, y^2 = 4 \cdot 2 \cdot y$

Comparing the above equation of parabola with the standard equation of parabola whose axis is the x-axis, we get,

$\rm a = 2$

So, the coordinates of the focus of the parabola are S(a, 0) = (2, 0).

Now, we use the distance formula to find the focal distance of Point P(2, 4) from focus S(2, 0). Let d be the focal distance, we get,

$\rm d = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\rm or, d = \sqrt{ (2 - 2)^2 + (4 - 0)^2 }$

$\rm or, d = \sqrt{ 0 + 4^2}$

$\rm or, d = \sqrt{4^2}$

$\rm \therefore d = 4$ units

Hence, the required focal distance of the point is found to be 4 units.

The focal distance of point P(4,1) is the distance from P to the focus of the parabola S(0,a).

The given equation of the parabola is $\rm x^2 = 16y$.

$\rm or, x^2 = 4 \cdot 4 y$

Comparing the above equation of parabola with the standard equation of parabola whose axis is the y-axis, we get,

$\rm a = 4$

So, the coordinates of the focus of the parabola are S(0, a) = (0,4).

Now, we use the distance formula to find the focal distance of Point P(4,1) from focus S(0,4). Let d be the focal distance, we get,

$\rm d = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\rm or, d = \sqrt{ (4 - 0)^2 + (1 - 4)^2}$

$\rm or, d = \sqrt{ 4^2 + (-3)^2}$

$\rm or, d = \sqrt{ 16 + 9}$

$\rm or, d = \sqrt{25}$

$\rm \therefore d = 5$ units.

Hence, the required focal distance of the point is found to be 5 units.

For a point P (x,y), x is the abscissa of P and y is the ordinate of P. It is given that the ordinate is two times that of abscissa. We can write it as $\rm y = 2x$.

Given equation of parabola is $\rm y^2 =36x$

Put the value of $\rm y = 2x$ in the equation of parabola and solve, we get,

$\rm (2x)^2 = 36x$

$\rm or, 4x^2 = 36x$

$\rm or, 4x^2 - 36x = 0$

$\rm or, 4 x ( x - 9) = 0$

Either

$\rm 4x = 0$

$\rm \therefore x = 0$

Or

$\rm x - 9 = 0$

$\rm \therefore x = 9$

For $\rm x = 0$, the value of the ordinate (y) is given by

$\rm y = 2 \cdot x = 2 \cdot 0 = 0$

For $\rm x = 9$, the value of the ordinate (y) is given by

$\rm y = 2 \cdot x = 2 \cdot 9 \ 18$

Hence, the required points on the parabola $\rm y^2 = 36x$ where the ordinate is two times that of its abscissa are (0,0) and (9,18). 

Let the vertex of the parabola $\rm x^2 = 12y$ be represented by the point $\rm C (h,k)$. The vertex of the given parabola is the origin. This means $\rm (h,k) = (0,0)$.

The given equation of the parabola is $\rm x^2 = 12y$

$\rm or, x^2 = 4 \cdot 3 \cdot y$

Comparing the above equation with $\rm x^2 = 4ay$, we get, $\rm a = 3$.

The coordinates of the focus of the parabola are $\rm S (0, a) = (0, 3)$.

Let the two ends of the latus rectum of the parabola be represented by the points P and Q. The y-coordinates of P and Q are the same as the y-coordinate of the vertex of the parabola.

Let the coordinates of P and Q be $\rm (x_1, y)$ and $\rm (x_2, y)$

From above, $\rm y = a = 3$

Let P and Q be the two endpoints of the latus rectum of the parabola. When we form a triangle by joining the points P, Q, and C, we get a triangle PQC whose height is equal to the distance of the focus from the vertex and whose base is equal to the length of the latus rectum.

So, area of the triangle PQC = $\rm \frac{1}{2} \cdot bh$

$\rm = \frac{1}{2} \cdot 4a \cdot a$

$\rm = 2a^2$

$\rm = 2 \cdot (3)^2$

$\rm = 2 \cdot 9$

$\rm = 18$ sq. units.

Hence, the required area of the triangle formed by joining the vertex and the two endpoints of the latus rectum of the parabola $\rm x^2 = 12y$ is 18 sq. units.

Any point (x,y) on the parabola $\rm y^2 = 4ax$ can be represented in parametric form as $\rm (x,y) = (at^2 , 2at)$.

Comparing the given equation of the parabola $\rm y^2 = 10x$ with $\rm y^2 = 4ax$, we get,

$\rm a = \frac{10}{4} = \frac{5}{2}$

Hence, $\rm (x, y) = \left ( \frac{5}{2} \cdot t^2 , 2 \cdot \frac{5}{2} \cdot t \right )$

$\rm \therefore (x,y) = \left ( \frac{5}{2} t^2, 5t  \right )$

Thus, the required parametric equation of the parabola is

$\rm x = \frac{5}{2} t^2$ and $\rm y = 5t$

We can represent any point (x,y) on the parabola $\rm x^2 = 4ay$ in the parametric form as $\rm (x, y) = (2at, at^2)$.

Comparing the given equation of the parabola $\rm x^2 = -16y$ with $\rm x^2 = 4ay$, we get, $\rm a = -4$.

Now, we solve for the parametric equation of the parabola

$\rm x = 2at$

$\rm or, x = 2 \cdot (-4) \cdot t$

$\rm \therefore x = -8t$

And,

$\rm y = at^2$

$\rm or, y = - 4 \cdot t^2$

$\rm \therefore y = -4t^2$

Hence, the required parametric equation of the parabola $\rm x^2 = -16y$ is $\rm x = -8t, y = -4 t^2$.

The parametric equation of the parabola $\rm ( y - k)^2 = 2 ( x - h)$ is $\rm (x,y) = (h + at^2, k + 2at)$.

We compare the given equation of the parabola $\rm (y - 2)^2 = 2 (x + 3)$ with $\rm ( y - k)^2 = 2 ( x -h)$, we get,

$\rm (h,k) = (-3, 2)$ and $\rm a = \frac{2}{4} = \frac{1}{2}$

Now, we solve for the value of x,

$\rm x = h + at^2$

$\rm or, x = -3 + \frac{1}{2} t^2$

And, we solve for the value of y,

$\rm y = k + 2at$

$\rm or, y = 2 + 2 \cdot \frac{1}{2} \cdot t$

$\rm \therefore y = 2 + t$

Hence, the required parametric equation of the parabola $\rm ( y - 2)^2 = 2 (x  + 3)$ is $\rm x = -3 + \frac{1}{2} t^2$ and $\rm y = 2 + t$.