Question

The line $x-y=1$ intersects the parabola $y^2=4x$ at $A$ and $B$. Normals at $A$ and $B$ intersect at $C$. If $D$ is the point at which line $CD$ is normal to the parabola, then coordinates of $D$ are

• A. $(4,-4)$
• B. $(4,4)$
• C. $(-4,-4)$
• D. none of these

Answer 1

The correct option is A $(4,-4)$

Given equation of line is
$y=x-1$
Given equation of parabola is
$y^2=4x$
$\Rightarrow (x-1{)}^2=4x$
$\Rightarrow x^2-6x+1=0$
$\Rightarrow x=3\pm \sqrt{8}$
$\therefore y=2\pm \sqrt{8}$
So, the line intersects the parabola at points $A(3+\sqrt{8},2+\sqrt{8})$ and $B(3-\sqrt{8},2-\sqrt{8})$
Suppose point D is $(x_3,y_3)$, then the sum of ordinates of the feets of normals drawn from point C is 0.
$y_1+y_2+y_3=0$
$\Rightarrow 2+\sqrt{8}+2-\sqrt{8}+y_3=0$
$\Rightarrow y_3=-4$
$\Rightarrow x_3=4$
Therefore, the point is $(4,-4)$.