In this page, you can find the complete solutions of the first exercise of Conic Section - Parabola chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, circle is the 12th chapter and has two exercises only. Out of which, this is the solution of the first exercise.


Check: Basic Mathematics Grade 11 (Sukunda Publication) Guide:

Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal


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Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here.


Do Question Number 3 Yourself. In case you can't do it, let us know in the comments section.



















4 a) Find the point on the parabola $y^2 = 8x$ at which the ordinate is double the abscissa.

Solution:
Let a be the abscissa and b be the ordinate. We have the coordinate (a,b) where b = 2a.

Equation of parabola: $y^2 = 8x$

Solving equation of parabola:
$(b)^2 = 8(a)$
$or, (2a)^2 = 8a$
$or, 4a^2 = 8a$
$or, 4a^2 - 8a = 0$
$or, 4a(a - 2) = 0$
$\implies \text{Either} \ a = 0 \ \text{or}\ a = 2$

Again,
$y^2 = 8x$
$or, b^2 = 8 * a$
When a = 0
$or, b^2 = 8 * 0$
$\therefore b = 0$
When a = 2
$or, b^2 = 8*2$
$or, b^2 = 16$
$\therefore b = \pm 4$
Hence, the required coordinates (a,b) are (0,0), (2,4), (2,-4).


4 b) Find the point on the parabola $x^2 = 12y$ at which the abscissa is twice the ordinate.

Solution:
Let a be the abscissa and b be the ordinate. We have the coordinate (a,b) where a = 2b.

Equation of parabola: $x^2 = 12y$

Solving equation of parabola:

$(a)^2 = 12 * b$
$or, (2b)^2 = 12 b$
$or, 4b^2 - 12b = 0$
$or, 4b(b - 3) = 0$
$\implies \text{Either} \  b = 0 \ \text{or}\  b = 3$

Again,
$x^2 = 12y$
$or, a^2 = 12* b$
When b = 0
$or,a^2 = 12 * 0$
$\therefore a = 0$
When b = 3
$or, a^2 = 12*3$
$or, a^2 = 36$
$\therefore b = \pm 6$
Hence, the required coordinates (a,b) are (0,0), (6,3), (-6,3).


5 a) Find the equation of the parabola with vertex at the origin, axis parallel to x-axis and passing through the point (1,-3).

Solution:
According to the question,
vertex is at the origin and axis is parallel to x-axis.

This means the equation of parabola is $y^2 = 4ax$

Given, passing point (x,y) = (1,-3)

Put the values of (x,y) in equation of parabola, we get,

$or, (-3)^2 = 4 a (1)$
$\therefore a = \dfrac{9}{4}$
Now,
equation of parabola is
$y^ 2 = 4ax$
$or, y^2 = 4* \dfrac{9}{4} x$
$\therefore y^2 = 9x$ is the required equation.


5 b) Find the equation of the parabola with focal width 16, axis parallel to x-axis and passing through the points (3,7) and (3,-1).

Solution:

Here,
focal width (4a) = 16
Axis is parallel to x-axis. This means equation of parabola is

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

$(y - k)^2 = 16 (x - h)$

Given,
Passing points are (3,7) and (3,-1).

When (x,y) = (3,7) in given parabola,

$(7 - k)^2 = 16 (3 - h)$ - (i)

When (x,y) = (3,-1) in given parabola,

$(-1 - k)^2 = 16 ( 3 - h)$ - (ii)

Equating the obtained two equation (i) and equation (ii), we get,

$(7 - k)^2 = (-1-k)^2$
$or, 49 - 14k = 1 + 2k$
$or, 48 = 16k$
$\therefore k = 3$
Put value of k = 3 in equation (ii), we get,

$or, (-1-3)^2 = 16(3- h)$
$or, 16 = 16 (3 -h)$

$or, 1 = 3 - h$

$\therefore h = 2$
So, required values of (h,k) = (2,3)

Hence, the required equation of parabola is $(y - 3)^2 = 16 (x -2)$


5 c) Obtain the equation of the parabola with axis parallel to y-axis, length of latus rectum 8 and passing through the points (2,1) and (-2,1).

Solution:

Here,
length of latus rectum (4a) = 8

Axis is parallel to y-axis. This means equation of parabola is

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

$(x-h)^2 = 8 (y - k)$

Given,
Passing points are (2,1) and (-2,1).

When (x,y) = (2,1) in given parabola,

$(2-h)^2 = 8 (1-k)$ - (i)
When (x,y) = (-2,1) in given parabola,

$(-2-h)^2 = 8 ( 1-k)$ - (ii)

Equating the obtained two equation (i) and equation (ii), we get,

$(2 - h)^2 = (-2-h)^2$
$or, -4h + 4 = 4h + 4$
$or, 8h = 0$
$\therefore h= 0$
Put value of h = 0 in equation (ii), we get,

$or, (-2  -0)^2 = 8 (1 - k)$

$or, \dfrac{1}{2} = 1 -k$
$or, k = 1- \dfrac{1}{2}$
$\therefore k = \dfrac{1}{2}$

So, required values of (h,k) = (0,$\frac{1}{2}$)

Hence, the required equation of parabola is

$(x - 0)^2 = 8 ( y - \frac{1}{2})$

$\implies x^2 = 8y - 4$


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About the Textbook:

Name: Basic Mathematics Grade XI
Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya
Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)
Telephone: 5320379, 5353537
Price: 695 /- (2078 BS)