Question

The angle between the pair of tangents of the parabola $y^2+12x=0$ which are normal to $x^2+y^2-6x-7y-4=0$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is D $\frac{\pi }{2}$

Given parabola $y^2+12x=0$
$\Rightarrow y^2=-12xa=-3$
Equation of tangent in slope form
$y=mx+\frac{a}{m}\Rightarrow y=mx-\frac{3}{m}$
This is a normal to the circle $x^2+y^2-6x-7y-4=0$, so it passes through the center of the circle $(3,\frac{7}{2})$
$\frac{7}{2}=3m-\frac{3}{m}\Rightarrow 6m^2-7m-6=0$
Roots of the above equation are $m_1,m_2$
Angle between the tangents
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
As $m_1{m}_2=-1$, so
$\theta =\frac{\pi }{2}$


Alternate solution:
Given parabola $y^2+12x=0$
$\Rightarrow y^2=-12xa=-3$
Equation of directrix is
$x=-a\Rightarrow x=3$
Center of $x^2+y^2-6x-7y-4=0$ is $(3,\frac{7}{2})$
Center lies on the directrix, so the angle between pair of tangents drawn from directrix to parabola is $\frac{\pi }{2}$