Let the eccentricity of the ellipse x^2 + a^2y^2 = 25a^2 be b times the eccentricity of th…

Question

Let the eccentricity of the ellipse ${x^2} + {a^2}{y^2} = 25{a^2}$ be b times the eccentricity of the hyperbola ${x^2} - {a^2}{y^2} = 5$, where a is the minimum distance between the curves y = ex and y = logex. Then $${a^2} + {1 \over {{b^2}}}$$ is equal to :

Accepted Answer

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

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

Solutions · 1