Question

The equation of the other normal to the parabola $y^2=4ax$ which passes through the intersection of those at $(4a,-4a)$ and $(9a,-6a)$ is:

• A. $5x-y+115a=0$
• B. $5x+y-135a=0$
• C. $5x-y-115a=0$
• D. $5x+y+115=0$

Answer 1

The correct option is B $5x+y-135a=0$

There is a slight misprint and the equation of the parabola should be $y^2=4ax$.

Any point on the parabola is of the form P ($a{t}^2$,$2at$).

Now, differentiating the equation of the parabola,

$2y\frac{dy}{dx}=4a\Rightarrow \frac{dy}{dx}=\frac{2a}{y}$

So, Slope of the Normal$=\frac{-y}{2a}=\frac{-2at}{2a}=-t$

Hence, Equation of Normal at P ($a{t}^2$,$2at$) is:

$y-2at=-t(x-a{t}^2)$

$\Rightarrow a{t}^3+2at-xt-y=0$ ...(1)

In this cubic equation, sum of roots =0

Hence, we can say $t_1+t_2+t_3=0$

Now, at point 1 ($4a$,$-4a$),

$-4a=2a{t}_1$

$\Rightarrow t_1=-2$

Similarly, at point 2 ($9a$,$-6a$),

$2a{t}_2=-6a$

$\Rightarrow t_2=-3$

Now as $t_1+t_2+t_3=0$,

we get, $t_3=5$

Putting $t_3=5$,

From the equation of normal in (1) above, we get

$a\left(125\right)+5(-x+2a)-y=0$

$\Rightarrow 135a=5x+y$

or, $5x+y-135a=0$ ....(B)