JEE PYQ: Parabola Question 19
Question 19 - 2019 (12 Apr Shift 2)
The equation of a common tangent to the curves, $y^2 = 16x$ and $xy = -4$, is:
(1) $x - y + 4 = 0$
(2) $x + y + 4 = 0$
(3) $x - 2y + 16 = 0$
(4) $2x - y + 2 = 0$
Show Answer
Answer: (1)
Solution
Tangent to $y^2 = 16x$: $y = mx + \frac{4}{m}$. Substituting in $xy = -4$: $x\left(mx + \frac{4}{m}\right) + 4 = 0 \Rightarrow mx^2 + \frac{4}{m}x + 4 = 0$. $D = 0$: $\frac{16}{m^2} - 16m = 0 \Rightarrow m^3 = 1 \Rightarrow m = 1$. Tangent: $y = x + 4$.
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