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Q.

Let $L_{1}$ be a tangent to the parabola $y^{2} = 4 (x + 1)$ and L₂ be a tangent to the parabola $y^{2} = 8 (x + 2)$ such that L₁ and L₂ intersect at right angles. Then L₁ and L₂meet on the straight line

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a

$x + 3 = 0$

b

$x + 2 y = 0$

c

$x + 2 = 0$

d

$2 x + 1 = 0$

answer is A.

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Detailed Solution

Tangent of $y^{2} = 4 (x + 1) is L_{1} : y = m (x + 1) + \frac{1}{m}$

and tangent to
$y^{2} = 8 (x + 2) is L_{2} : y = \frac{- 1}{m} (x + 2) - 2 m$
Equating both $m (x + 1) + \frac{1}{m} = \frac{- 1}{m} (x + 2) - 2 m \Rightarrow x (m + \frac{1}{m}) + m + 2 m + \frac{1}{m} + \frac{2}{m} = 0 \Rightarrow x (m + \frac{1}{m}) + 3 (m + \frac{1}{m}) = 0 ∴ x + 3 = 0$

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Let L1be a tangent to the parabola y2=4x+1 and L2 be a tangent to the parabola y2=8x+2 such that L1 and L2 intersect at right angles. Then L1 and L2 meet on the straight line