Mathematics · 2022
JEE Main · 28 July 2022, Shift 1 · Q27
For the hyperbola H: x^2-y^2=1 and the ellipse E: x^2/(a^2)+y^2/(b^2)=1, a > b >0, let the (1) eccentricity of E be reciprocal of the eccentricity of…
For the hyperbola $\displaystyle \mathrm{H}: x^2-y^2=1$ and the ellipse $\displaystyle \mathrm{E}: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}>0$, let the
(1)
eccentricity of E be reciprocal of the eccentricity of H , and
(2)
the line $\displaystyle y=\sqrt{\frac{5}{2}} x+\mathrm{K}$ be a common tangent of E and H.
Then $\displaystyle 4\left(\mathrm{a}^2+\mathrm{b}^2\right)$ is equal to \_\_\_\_.
Official answer
3
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.