JEE Main 2021 (Online) 22nd July Evening ShiftQuestion 1
Let $E_1:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, $a>b$. Let $E_2$ be another ellipse such that it touches the end points of major axis of $E_1$ and the foci of $E_2$ are the end points of minor axis of $E_1$. If $E_1$ and $E_2$ have same eccentricities, then its value is :
Ellipse
Solution
Concept. $E_2$ has foci $(0,\pm b)$, so its major axis is the $y$-axis; it passes through $(\pm a,0)$, so its semi-minor axis is $a$. Formulas used. (1) Vertical ellipse $\frac{x^2}{p^2}+\frac{y^2}{q^2}=1$ ($q>p$): foci $(0,\pm c)$, $c^2=q^2-p^2$, $e^2=\frac{c^2}{q^2}$(2) $E_1$: $e^2=1-\frac{b^2}{a^2}$Step 1 - $E_2$ (1). $p=a$, $c=b$, so $q^2=a^2+b^2$ and $e_2^2=\dfrac{b^2}{a^2+b^2}$.Step 2 - Equal eccentricities. Let $t=\frac{b^2}{a^2}$. Using (2): $$\frac{t}{1+t}=1-t\ \Rightarrow\ t=1-t^2\ \Rightarrow\ t^2+t-1=0$$ $t>0$, so $t=\frac{\sqrt5-1}2$. Step 3 - $e$. $e^2=1-t=t^2$ (from $t^2+t-1=0$), so $e=t=\dfrac{-1+\sqrt5}{2}$. Answer: A