Question

The normal at $\left( {a,2a} \right)$ on ${y^2} = 4ax$ meets the curve again at $\left( {a{t^2},2at} \right)$ . Then the value of its parameter is equal to
A. 1
B. 3
C. -1
D. -3

Answer 1

Hint : First find the slope of the tangent at point $\left( {a,2a} \right)$ by differentiating the equation ${y^2} = 4ax$ with respect to x. The slopes of perpendicular lines are negative reciprocals of one another. So using this, find the slope of normal at point $\left( {a,2a} \right)$ as normal and tangent are perpendicular to each other. Using the slope of the normal line and point $\left( {a,2a} \right)$ , find the line equation of the normal line. Substitute the line equation in the curve equation ${y^2} = 4ax$ to find the parameter t.
Formulas used:
Slope-point form of a line equation is $\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$ , where m is the slope and $\left( {{x_1},y1} \right)$ is the given point.

Complete step by step solution:
We are given that the normal at $\left( {a,2a} \right)$ on ${y^2} = 4ax$ meets the curve again at $\left( {a{t^2},2at} \right)$ .
We have to find the value of t.
First we have to find the slope of tangent at point $\left( {a,2a} \right)$ .

So we are differentiating ${y^2} = 4ax$ with respect to x.
We get,
$\dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( {4ax} \right)$
$\Rightarrow 2y\dfrac{{dy}}{{dx}} = 4a\left( {\dfrac{{dx}}{{dx}}} \right)$
Let $\dfrac{{dy}}{{dx}}$ be $y'$ .
$\Rightarrow 2yy' = 4a\left( {\because \dfrac{{dx}}{{dx}} = 1} \right)$
$\Rightarrow y' = \dfrac{{4a}}{{2y}}$
Slope of tangent at point $\left( {a,2a} \right)$ is $\dfrac{{4a}}{{2\left( {2a} \right)}} = \dfrac{{4a}}{{4a}} = 1$ as y-coordinate of $\left( {a,2a} \right)$ is 2a.
Tangent and normal are perpendicular lines. So the slope of the normal line at point $\left( {a,2a} \right)$ is $\dfrac{{ - 1}}{{\left( 1 \right)}} = - 1$
We have got the slope of the normal and the point, $\left( {a,2a} \right)$ , from which the normal travels.
Therefore, the equation of the normal line is $\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$
$\Rightarrow y - 2a = - 1\left( {x - a} \right)$
$\Rightarrow y - 2a = - x + a$
$\Rightarrow y = 3a - x$
On substituting the value of y as $3a - x$ in ${y^2} = 4ax$ , we get
$\Rightarrow {\left( {3a - x} \right)^2} = 4ax$
$\Rightarrow 9{a^2} + {x^2} - 6ax = 4ax$
$\Rightarrow {x^2} - 6ax - 4ax + 9{a^2} = 0$
$\Rightarrow {x^2} - 10ax + 9{a^2} = 0$
We are next finding the factors of the above equation
$\Rightarrow {x^2} - ax - 9ax + 9{a^2} = 0$
$\Rightarrow x\left( {x - a} \right) - 9a\left( {x - a} \right) = 0$
$\Rightarrow \left( {x - a} \right)\left( {x - 9a} \right) = 0$
$\therefore x = a,x = 9a$
The value of x is 9a as we already know when x is a, y is 2a in the point $\left( {a,2a} \right)$
This gives,
$y = 3a - x = 3a - 9a = - 6a$
Therefore, the x and y coordinates of point $\left( {a{t^2},2at} \right)$ are 9a and -6a respectively.
This means that
$\Rightarrow 2at = - 6a$
$\therefore t = \dfrac{{ - 6a}}{{2a}} = - 3$
Therefore, the parameter t is equal to -3.
So, the correct answer is option (D), “ -3”.

Note : Here we have a slope and point given so we have used slope-point form to find the line equation of the normal. If two points of a line are given, then we have to use two-point form to find the line equation. A tangent is a straight line to a plane curve at a given point that just touches the curve only at one (given) point. Do not confuse a tangent with a secant.