JEE PYQ: Parabola Question 18
Question 18 - 2019 (09 Apr Shift 2)
If the tangent to the parabola $y^2 = x$ at a point $(\alpha, \beta)$, $(\beta > 0)$ is also a tangent to the ellipse, $x^2 + 2y^2 = 1$, then $\alpha$ is equal to:
(1) $\sqrt{2} - 1$
(2) $2\sqrt{2} - 1$
(3) $2\sqrt{2} + 1$
(4) $\sqrt{2} + 1$
Show Answer
Answer: (4)
Solution
Tangent to $y^2 = x$ at $\left(\frac{1}{4m^2}, -\frac{1}{2m}\right)$: $y = mx \pm \sqrt{m^2 + \frac{1}{2}}$. For common tangency: $\frac{1}{4m} = \sqrt{m^2 + \frac{1}{2}}$. Solving: $16m^4 + 8m^2 - 1 = 0$, $m^2 = \frac{\sqrt{2}-1}{4}$. Then $\alpha = \frac{1}{4m^2} = \frac{1}{\sqrt{2}-1} = \sqrt{2}+1$.
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