Grade 11 Parabola and a Line (Conic Section) Exercise 2 Solutions | Basic Mathematics Grade XI by Sukunda Pustak Bhawan


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

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


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

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


Disclaimer:

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.

5 a) Find the equation of the (i) tangent (ii) normal, to the parabola $y^2 = 6x$ making angle 45$^o$ with the x-axis.

Solution:
Equation of parabola is $y^2 = 6x$

Comparing equation of parabola with standard equation of parabola $y^2 = 4ax$, we get, $a =  \dfrac{6}{4} = \dfrac{3}{2}$

The tangent makes angle 45$^o$ with the X-axis $\implies \theta = 45^o$

So, slope of tangent (m) = $tan \theta$ = $tan 45^o$ = $1$

$\implies m = 1$
(i) Equation of tangent to the parabola is

$y = mx + \dfrac{a}{m}$
$or, y = 1x + \dfrac{ \frac{3}{2}}{1}$

$or, y = x + \dfrac{3}{2}$
$or, 2y = 2x + 3$
$\therefore 2x - 2y + 3 = 0$
(ii) Equation of normal to the parabola is

We know, tangent and normal are perpendicular to each other. Therefore, slope of normal = - 1 / slope of tangent.

$\implies \text{slope of normal} (m_1} = - \dfrac{1}{m}$

$\implies m_1 = -1$
So,
$y = m_1x - 2am_1 - am_1^3$
$or, y = (-1) x - 2 * \dfrac{3}{2} * (-1) - \dfrac{3}{2} * (-1)^3$

$or, y = -x+ 3 + dfrac{3}{2}$
$or, y = \dfrac{-2x + 6 + 3}{2}$

$or, 2y = -2x + 9$
$\therefore 2x + 2y - 9 = 0$
6 b) Prove that the tangent at the extrincities of the latus rectum of a parabola $y^2 = 16x$ are at right angle.

Solution:
From the equation of the parabola $y^2 = 16x$, we get to know that S(a,0) = (4,0).

We know,
Co-ordinates of points at the extrincities of latus rectum are $(a,\pm 2a)$

Let the point above X-axis be P1 and that below the X-axis be P2 then,

$P_1 = (4,8)$ and $P_2 = (4,-8)$

Equation of tangent to parabola is

$y = mx + \dfrac{a}{m}$ - (i)
where m is the slope of the tangent.

Let $m_1$ and $m_2$ be the slopes of the two tangents.

Putting values of (x,y) from P1 in equation (i), we get,

$8 = 4m_1 + \dfrac{4}{m_1}$
On solving, we get,
$m_1 = 1$
Similarly, on putting values of (x,y) from P2 in equation (ii), wegget,

$-8 = 4m_2 + \dfrac{4}{m_2}$
$m_2 = -1$
Since, product of slopes of two tangents is -1, the tangents are at right angle.

i.e., $m_1 * m_2 = (1)*(-1) = (-1)$

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)