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Let L1 be a t...

Question

Let $L_{1}$ be a tangent to the parabola $y^{2} = 4 (x + 1)$ and $L_{2}$ be a tangent to the parabola $y^{2} = 8 (x + 2)$ such that $L_{1}$ and $L_{2}$ intersect at right angles. Then $L_{1}$ and $L_{2}$ meet on the straight line:

A

x + 2 y = 0

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B

x + 2 = 0

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C

2 x + 1 = 0

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D

x + 3 = 0

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Solution

The correct option is D $x + 3 = 0$
The equation of the tangent to the parabola $y^{2} = 4 a x$ at $(a t^{2}, 2 a t)$ is given by $t y = x + a t^{2}$
Given that $L_{1}$ is the tangent of $y^{2} = 4 (x + 1)$
$L_{1} : t_{1} y = (x + 1) + t_{1}^{2} \dots (1)$
and $L_{2}$ is the tangent of $y^{2} = 8 (x + 2)$
$L_{2} : t_{2} y = (x + 2) + 2 t_{2}^{2} \dots (2)$
$L_{1}$ is perpendicular to $L_{2}$
$\frac{1}{t_{1}} \cdot \frac{1}{t_{2}} = - 1$
$t_{1} t_{2} = - 1$
$t_{2} (1) - t_{1} (2)$
$t_{1} t_{2} y = t_{2} (x + 1) + t_{2} t_{1}^{2}$
$t_{1} t_{2} y = t_{1} (x + 2) + 2. t_{1}^{2} t_{1}$
Subtracting the above equations, we get
$(t_{2} - t_{1}) x + (t_{2} - 2 t_{1}) + t_{2} t_{1} (t_{1} - 2 t_{2}) = 0$
$(t_{2} - t_{1}) x + (t_{2} - 2 t_{1}) - (t_{1} - 2 t_{2}) = 0$
$(t_{2} - t_{1}) x + 3 t_{2} - 3 t_{1} = 0$
$\Rightarrow x + 3 = 0$

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