JEE PYQ: Application Of Derivatives Question 39
Question 39 - 2020 (06 Sep Shift 1)
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:
(1) $x + 3 = 0$
(2) $2x + 1 = 0$
(3) $x + 2 = 0$
(4) $x + 2y = 0$
Show Answer
Answer: (1)
Solution
Tangent to $y^2 = 4(x+1)$: $y = m(x+1) + \frac{1}{m}$. Tangent to $y^2 = 8(x+2)$: $y = m’(x+2) + \frac{2}{m’}$ with $m’ = -\frac{1}{m}$. Setting equal and solving: locus is $x + 3 = 0$.
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