Question

Consider the parabola whose focus at $(0,0)$ and tangent at vertex is $x-y+1=0$.
Tangents drawn to the parabola at the extremities of the chord $3x+2y=0$ intersects at an angle

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

Answer 1

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


The distance between the focus and the tangent at the vertex is $\frac{|0-0+1|}{\sqrt{1^2+1^2}}=\frac{1}{\sqrt{2}}$
The directrix is the line parallel to the tangent at vertex and at a distance $2\times \frac{1}{\sqrt{2}}$ from the focus.
Let the equation of the directirx be,
$x-y+\lambda =0$
So,
$\left|\frac{\lambda }{\sqrt{1^2+1^2}}\right|=\frac{2}{\sqrt{2}}$
$\Rightarrow \lambda =2$
Let $P(x,y)$ be any moving point on the parabola. Then,
$OP=PM$
$x^2+y^2={\left(\frac{x-y+2}{\sqrt{1^2+1^2}}\right)}^2$
$\Rightarrow 2x^2+2y^2=(x-y+2{)}^2$
$\Rightarrow x^2+y^2+2xy-4x+4y-4=0$
Latus rectum length
$=2\times$ (Distance of focus from directrix)
$=2\left|\begin{array}{c}\frac{0-0+2}{\sqrt{1^2+1^2}}\end{array}\right|=2\sqrt{2}$
Solving the parabola with the x-axis,
$x^2-4x-4=0$
$\Rightarrow x=\frac{4\pm \sqrt{32}}{2}=2\pm 2\sqrt{2}$
Therefore, the length of chord on the x-axis is $4\sqrt{2}$
Since the chord $3x+2y=0$ passes through the focus, it is focal chord.
Hence, tangents at the extremities of chord are perpendicular.