JEE Advanced 2020 Paper 1Question 1
Let $a,b$ and $\lambda$ be positive real numbers. Suppose $P$ is an end point of the latus rectum of the parabola $y^2=4\lambda x$, and suppose the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through the point $P$. If the tangents to the parabola and the ellipse at the point $P$ are perpendicular to each other, then the eccentricity of the ellipse is
Ellipse
Solution
Concept. Find both tangent slopes at $P=(\lambda,2\lambda)$ by differentiation; their product is $-1$. This fixes $b^2:a^2$. Formulas used. (1) End of latus rectum of $y^2=4\lambda x$: $(\lambda,\pm2\lambda)$(2) Slope of tangent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $(x,y)$: $-\frac{b^2x}{a^2y}$(3) If $b>a$ the major axis is along $y$: $e^2=1-\frac{a^2}{b^2}$Step 1 - Point $P$. Take $P=(\lambda,2\lambda)$ (the other end gives the same result by symmetry). Step 2 - Parabola slope. $2yy'=4\lambda\Rightarrow y'=\frac{2\lambda}{y}=\frac{2\lambda}{2\lambda}=1$. Step 3 - Ellipse slope by (2). $m=-\frac{b^2\cdot\lambda}{a^2\cdot2\lambda}=-\frac{b^2}{2a^2}$. Step 4 - Perpendicular. $1\cdot\Big(-\frac{b^2}{2a^2}\Big)=-1\Rightarrow b^2=2a^2$. Step 5 - Eccentricity. $b>a$, so by (3): $$e^2=1-\frac{a^2}{2a^2}=\frac12\Rightarrow e=\frac1{\sqrt2}$$(Also $P$ on the ellipse gives $\frac{\lambda^2}{a^2}+\frac{4\lambda^2}{2a^2}=1$, i.e. $a^2=3\lambda^2$; the figure uses $\lambda=1$.) Answer: A