For the hyperbola $\mathrm{H}: x^{2}-y^{2}=1$ and the ellipse $\mathrm{E}: \frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1$, a $>\mathrm{b}>0$, let the

(1) eccentricity of $\mathrm{E}$ be reciprocal of the eccentricity of $\mathrm{H}$, and

(2) the line $y=\sqrt{\frac{5}{2}} x+\mathrm{K}$ be a common tangent of $\mathrm{E}$ and $\mathrm{H}$.

Then $4\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)$ is equal to _____________.

<p>The equation of tangent to hyperbola ${x^2} - {y^2} = 1$ within slope $m$ is equal to $y = mx\, \pm \,\sqrt {{m^2} - 1}$ ...... (i)</p> <p>And for same slope $m$, equation of tangent to ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ is $y = mx\, \pm \,\sqrt {{a^2}{m^2} + {b^2}}$ ...... (ii)</p> <p>$\because$ Equation (i) and (ii) are identical</p> <p>$\therefore$ ${a^2}{m^2} + {b^2} = {m^2} - 1$</p> <p>$\therefore$ ${m^2} = {{1 + {b^2}} \over {1 - {a^2}}}$</p> <p>But equation of common tangent is $y = \sqrt {{5 \over 2}} x + k$</p> <p>$\therefore$ $$m = \sqrt {{5 \over 2}} \Rightarrow {5 \over 2} = {{1 + {b^2}} \over {1 - {a^2}}}$$</p> <p>$\therefore$ $5{a^2} + 2{b^2} = 3$ ....... (i)</p> <p>eccentricity of ellipse $= {1 \over {\sqrt 2 }}$</p> <p>$\therefore$ $1 - {{{b^2}} \over {{a^2}}} = {1 \over 2}$</p> <p>$\Rightarrow {a^2} = 2{b^2}$ ....... (ii)</p> <p>From equation (i) and (ii) : ${a^2} = {1 \over 2},\,{b^2} = {1 \over 4}$</p> <p>$\therefore$ $4({a^2} + {b^2}) = 3$</p>