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Definition of Functions

The normal at...

Question

The normal at $(a, 2 a)$ on $y^{2} = 4 a x$ meets the curve again at $(a t^{2}, 2 a t)$ . Then the value of its parameter is equal to

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Solution

The correct option is C $- 3$
$y^{2} = 4 a x$
$2 y . y^{'} = 4 a$
$y^{'} = \frac{2 a}{y}$
Hence, $y_{a, 2 a}^{'} = 1$
Hence slope of the normal will be -1.
Hence the equation of the normal will be
$\frac{y - 2 a}{x - a} = - 1$
$y - 2 a = - x + a$
$x + y = 3 a$
$y = 3 a - x$ ...(i)
$y^{2} = x^{2} - 6 a x + 9 a^{2}$
$4 a x = x^{2} - 6 a x + 9 a^{2}$
$x^{2} - 10 a x + 9 a^{2} = 0$
$(x - 9 a) (x - a) = 0$
$x = 9 a$ and $x = a$
Therefore $y = 3 a - x$
$y = - 6 a, 2 a$
Now, $2 a t = - 6 a$
$t = - 3$ .

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