Question

The normal at $P(2,4)$ to $y^2=8x$ meets the parabola at $Q.$ Then the equation of the circle having normal chord $PQ$ as diameter is

• A. $x^2+y^2-20x+8y-12=0$
• B. $x^2+y^2-10x+4y-8=0$
• C. $x^2+y^2-12x+6y-15=0$
• D. $x^2+y^2-10x+8y-12=0$

Answer 1

The correct option is A $x^2+y^2-20x+8y-12=0$

Normal at $({at}_1^2,2{at}_1)$ meets the parabola again at $({at}_2^2,2{at}_2)$
Then, $t_2=-t_1-\frac{2}{t_1}$
Given parabola is $y^2=8x$
$\Rightarrow a=2$
We are given $P=(2,4)$
$\Rightarrow 2{at}_1=4$
$\therefore t_1=1$
$\Rightarrow t_2=-t_1-\frac{2}{t_1}$
$\Rightarrow t_2=-3$
Then, $P=(2,4)$ and $Q=(18,-12)$
Then, equation of circle is given by $(x-2)(x-18)+(y-4)(y+12)=0$
$\therefore x^2+y^2-20x+8y-12=0$