Question

Two straight lines are perpendicular to each other, one of them touches the parabola $y^2=4a(x+a)$ and the other touches $y^2=4b(x+b)$. Their point of intersection lies on the line

• A. $x-a+b=0$
• B. $x+a-b=0$
• C. $x+a+b=0$
• D. $x-a-b=0$

Answer 1

The correct option is C $x+a+b=0$

Let slope of tangent to the parabola $y^2=4a(x+a)$ be $m.$
Then the slope of perpendicular line will be $-\frac{1}{m}$
Equation of tangent to the parabola $y^2=4a(x+a)$ is
$y=m(x+a)+\frac{a}{m}$ .....(1)
Equation of tangent to the parabola $y^2=4b(x+b)$ having slope $-\frac{1}{m}$is
$y=-\frac{1}{m}(x+b)+\frac{b}{-\frac{1}{m}}$
$\Rightarrow y=-\frac{1}{m}(x+b)-bm$ .....(2)
Subtracting (2) from (1), we get
$(m+\frac{1}{m})x+am+\frac{a}{m}+\frac{b}{m}+bm=0$
$(m+\frac{1}{m})x+(a+b)(m+\frac{1}{m})=0$
or $x+a+b=0$ which is a locus of their point of intersection.