Question

Let the normals at points $A\left(4 a , - 4 a\right)$ and $B\left(9 a , - 6 a\right)$ on the parabola $y^{2}=4ax$ meet at the point $P.$ The equation of the normal from $P$ on $y^{2}=4ax$ (other than $PA$ and $PB$ ) is

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

Answer 1

Answer: (A)

Let, $A=\left(a t_{1}^{2} , 2 a t_{1}\right), \, B=\left(a t_{2}^{2} , 2 a t_{2}\right)$
and the foot of the required normal is $\left(a t_{3}^{2} , 2 a t_{3}\right)$ , then
$2at_{1}=-4a, \, 2at_{2}=-6a$ and $t_{1}+t_{2}+t_{3}=0$
$\Rightarrow t_{1}=-2, \, t_{2}=-3$ and $-2-3+t_{3}=0$
$\Rightarrow t_{3}=5$
Hence, the equation of the required normal is $y=-t_{3}x+2at_{3}+at_{3}^{2}$
$\Rightarrow y=-5x+10a+125a$
$\Rightarrow y+5x-135a=0$