Two tangents are drawn to the parabola $y^{2} = 8 x$ which meets the tangent at vertex at $P$ and $Q$ respectively. If $P Q = 4 units$ , then the locus of the point of intersection of the two tangents is

y^{2} = 8 (x + 2)

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y^{2} = 8 (x - 2)

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x^{2} = 8 (y + 2)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} = 8 (y - 2)

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Solution

The correct option is A $y^{2} = 8 (x + 2)$
$S : y^{2} = 8 x \Rightarrow a = 2$
Let the two points on parabola be $R (t_{1})$ and $S (t_{2})$

Equation of the tangent at the vertex is
$x = 0$
Equation of tangent at $R (t_{1})$
$t_{1} y = x + a t_{1}^{2} \Rightarrow P \equiv (0, a t_{1})$
Similarly, we get
$Q \equiv (0, a t_{2})$
Now,
$\Rightarrow P Q = | a t_{1} - a t_{2} | = 4 \Rightarrow | t_{1} - t_{2} | = 2 (∵ a = 2) \Rightarrow (t_{1} - t_{2})^{2} = 4 \dots (i)$

Point of intersection of tangent at $R$ and $S$ is
$T (h, k) = (a t_{1} t_{2}, a (t_{1} + t_{2})) \Rightarrow t_{1} t_{2} = \frac{h}{a}, t_{1} + t_{2} = \frac{k}{a} \Rightarrow (t_{1} - t_{2})^{2} = (t_{1} + t_{2})^{2} - 4 t_{1} t_{2} \Rightarrow 4 = \frac{k^{2}}{a^{2}} - \frac{4 h}{a} \Rightarrow 4 = \frac{k^{2}}{4} - \frac{4 h}{2} (∵ a = 2)$

Hence, the locus is
$y^{2} = 8 (x + 2)$

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