Question

The locus of the point of intersection of the tangents drawn at the ends of a focal chord of the parabola $x^2=-8y$ is _______

• A. y = 2
• B. y = -2
• C. x = 2
• D. x = -2

Answer 1

Answer: (A)

Given that, parabola $x^2=-8y$

Here, on comparing with $x^2=4ay$
$\Rightarrow 4a=-8\Rightarrow a=-2$
Let the parametric coordinate of parabola
$x^2=-8y$ is, $P\to \left(4t,-2t^2\right)$
and the other coordinate of latusrectum is
$P^{\prime }\to \left(\frac{-4}{t},\frac{-2}{t^2}\right)$
Now, the equation tangent of parabola $x^2=-8y$
$x\cdot x_1=-4\left(y+y_1\right)$...(i)
At $Px(4t)=-4\left(y-2t^2\right)$
$xt=-y+2t^2$
$xt+y=2t^2$...(ii)
$\Rightarrow x\left(\frac{-4}{t}\right)=-4\left(y-\frac{2}{t^2}\right)$
$\Rightarrow \frac{x}{t}=y-\frac{2}{t^2}$
$\Rightarrow xt=y{t}^2-2$
$\Rightarrow xt-y{t}^2=-2$...(iii)
On solving Eqs. (i) and (ii)
$x{t}^3+y{t}^2=2t^4$
$xt-y{t}^2=-2$
___________________
$xt\left(1+t^2\right)=-2\left(1-t^4\right)$
$xt\left(1+t^2\right)=-2\left(1+t^2\right)\left(1-t^2\right)$
$tx=-2\left(1-t^2\right)$...(iv)
From Eq. (ii)
$-2\left(1-t^2\right)+y=2t^2$
$\Rightarrow -2+2t^2+y=2t^2$
$\Rightarrow y=2$
Hence, the intersection point of both tangent lying on $Q$.
ie, $y=2.$ Which is the required locus.