Question

Two straight lines are perpendicular to each other. One of them touches the parabola $y^2=4a(x+a)$ and the other touches $y^2=4b(x+b).$ Find the locus of point of intersection.

• A. $x+a+b=0$
• B. $x-a+b=1$
• C. $x+a-b=2$
• D. $x+a+b=2$

Answer 1

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

Any tangent to $y^2=4a(x+a)$ is
$y=m(x+a)+\frac{a}{m}......(Notex+a)$
Similarly $y=m^{\prime }(x+b)+b/m^{\prime }$ is a tangent to the other parabola. Since the tangents are perpendicular ${mm}^{\prime }=-1.$ In order to find the locus of their point of intersection we have to eliminate the two variables $m,m^{\prime }$ between the above relations.
Substracting, we have
$x(m-m^{\prime })+a(m+\frac{1}{m})-b(m+\frac{1}{m})=0$
Now put $m^{\prime }=-\frac{1}{m}$ and cancel $m+\frac{1}{m}.$
$\therefore x(m+\frac{1}{m})+a(m+\frac{1}{m})+b(m+\frac{1}{m})=0$
$\therefore x+a+b=0$ is the required locus.
Ans: A