QJEEM22Ma0539: JEE Main 2022 Mathematics - Class 11 JEEM22Ma0539 Solution: Let the eccentricity of the ellipse be b times the eccentricity …

Given ellipse ${x^2} + {a^2}{y^2} = 25{a^2} \Rightarrow {{{x^2}} \over {25{a^2}}} + {{{y^2}} \over {25}} = 1$

eccentricity $({e_1}) = \sqrt {1 - {{{b^2}} \over {{a^2}}}}$

$= \sqrt {1 - {{25} \over {25{a^2}}}}$

$= \sqrt {1 - {1 \over {{a^2}}}}$

$\Rightarrow e_1^2 = 1 - {1 \over {{a^2}}}$

Also, given hyperbola,

${x^2} - {a^2}{y^2} = 5$

$\Rightarrow {{{x^2}} \over 5} - {{{y^2}} \over {{5 \over {{a^2}}}}} = 1$

eccentricity $({e_2}) = \sqrt {1 + {{{b^2}} \over {{a^2}}}}$

$= \sqrt {1 + {5 \over {5{a^2}}}}$

$= \sqrt {1 + {1 \over {{a^2}}}}$

$\Rightarrow e_2^2 = 1 + {1 \over {{a^2}}}$

Also given,

${e_1} = b \times {e_2}$

$\Rightarrow e_1^2 = {b^2} \times e_2^2$

$\Rightarrow 1 - {1 \over {{a^2}}} = {b^2}\left( {1 + {1 \over {{a^2}}}} \right)$

$\Rightarrow {{{a^2} - 1} \over {{a^2}}} = {{{b^2}({a^2} + 1)} \over {{a^2}}}$

$\Rightarrow {b^2} = {{{a^2} - 1} \over {{a^2} + 1}}$

$y = \log _e^x$ is inverse of $y = {e^x}$ so it is mirror image of each other with respect to y = x line.

Slope of tangent to y = e^x curve

${{dy} \over {dx}} = {e^x}$

Slope of tangent to $y = \log _e^x$ curve,

${{dy} \over {dx}} = {1 \over x}$

Both tangents are parallel to y = x line for minimum distance condition.

$\therefore$ Slope of y = x line = Slope of both the tangent.

$\therefore$ ${{dy} \over {dx}} = {e^x} = 1 \Rightarrow {e^x} = {e^0} = x = 0$

$\therefore$ $y = {e^x} = {e^0} = 1$

and ${{dy} \over {dx}} = {1 \over x} = 1 \Rightarrow x = 1$

$\therefore$ $y = \log _e^1 = 0$

$\therefore$ tangent at (0, 1) point of $y = {e^x}$ curve and tangent at (1, 0) point of $y = \log _e^x$ curve are parallel.

$\therefore$ Minimum distance between point (0, 1) and (1, 0) is $= \sqrt {{1^2} + {1^2}} = \sqrt 2$

$\therefore$ $a = \sqrt 2$

So, ${b^2} = {{{a^2} - 1} \over {{a^2} + 1}} = {{2 - 1} \over {2 + 1}} = {1 \over 3}$

$\therefore$ ${a^2} + {1 \over {{b^2}}} = 2 + {1 \over {1/3}} = 5$