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 $x+a+b=0$

Any tangent to $y^2=4a(x+a)$ is,
$y=m(x+a)+\frac{a}{m}\cdots \left(i\right)$
Any tangent to $y^2=4b(x+b)$ which is perpendicular to (i) is
$y=-\frac{1}{m}(x+b)-bm\cdots \left(ii\right)$
Subtracting, we get
$(m+\frac{1}{m})x+(a+b)(m+\frac{1}{m})=0$
$\Rightarrow x+a+b=0$
Which is the locus of their point of intersection.