JEE PYQ: Parabola Question 26
Question 26 - 2019 (11 Jan Shift 1)
Equation of a common tangent to the parabola $y^2 = 4x$ and the hyperbola $xy = 2$ is:
(1) $x + y + 1 = 0$
(2) $x - 2y + 4 = 0$
(3) $x + 2y + 4 = 0$
(4) $4x + 2y + 1 = 0$
Show Answer
Answer: (3)
Solution
Tangent to $y^2 = 4x$: $y = mx + \frac{1}{m}$. For tangency with $xy = 2$: $x(mx + \frac{1}{m}) = 2 \Rightarrow mx^2 + \frac{x}{m} - 2 = 0$. $D = 0$: $\frac{1}{m^2} + 8m = 0 \Rightarrow 1 + 8m^3 = 0 \Rightarrow m = -\frac{1}{2}$. Tangent: $y = -\frac{1}{2}x - 2 \Rightarrow x + 2y + 4 = 0$.
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