Let the foci and length of the latus rectum of an ellipse x^2 a^2+ y^2 b^2=1, a>b be (± 5,0) and √50, respectively. Then, the square of the…

Given,

Ellipse $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$a2x2​+b2y2​=1

And its focii $\equiv (\pm 5,0)$≡(±5,0) and its latusrectum $\frac{2{b}^{2}}{a}=\sqrt{50}$a2b2​=50​

So, $ae=5$ae=5 and ${b}^{2}=\frac{5\sqrt{2}a}{2}$b2=252​a​

Now, using the formula eccentricity we get,

$${b}^{2}={a}^{2}(1-{e}^{2})=\frac{5\sqrt{2}a}{2}$$b2=a2(1−e2)=252​a​

$\Rightarrow a(1-{e}^{2})=\frac{5\sqrt{2}}{2}$⇒a(1−e2)=252​​ ${\text{as}a\neq 0}$asa=0

$$\Rightarrow \frac{5}{e}(1-{e}^{2})=\frac{5\sqrt{2}}{2}$$⇒e5​(1−e2)=252​​

$$\Rightarrow \sqrt{2}-\sqrt{2}{e}^{2}=e$$⇒2​−2​e2=e

$$\Rightarrow \sqrt{2}{e}^{2}+e-\sqrt{2}=0$$⇒2​e2+e−2​=0

$$\Rightarrow \sqrt{2}{e}^{2}+2e-e-\sqrt{2}=0$$⇒2​e2+2e−e−2​=0

$$\Rightarrow \sqrt{2}e(e+\sqrt{2})-1(1+\sqrt{2})=0$$⇒2​e(e+2​)−1(1+2​)=0

$$\Rightarrow (e+\sqrt{2})(\sqrt{2}e-1)=0$$⇒(e+2​)(2​e−1)=0

$$\therefore e\neq -\sqrt{2};e=\frac{1}{\sqrt{2}}$$∴e=−2​;e=2​1​

$\Rightarrow a=5\sqrt{2}$⇒a=52​ and $b=5$b=5

Now, finding eccentricity of hyperbola,

$$\frac{{x}^{2}}{{b}^{2}}-\frac{{y}^{2}}{{a}^{2}{b}^{2}}=1$$b2x2​−a2b2y2​=1

We know that formula of eccentricity of hyperbola is given by, ${B}^{2}={A}^{2}({{e}_{H}}^{2}-1)$B2=A2(eH​2−1) where {A}^{2}={b}^{2}&{B}^{2}={a}^{2}{b}^{2}

$$\Rightarrow {a}^{2}{b}^{2}={b}^{2}({e}_{H}^{2}-1)$$⇒a2b2=b2(eH2​−1)

$$\Rightarrow {a}^{2}=({e}_{H}^{2}-1)$$⇒a2=(eH2​−1)

$$\Rightarrow 50=({e}_{H}^{2}-1)$$⇒50=(eH2​−1)

$$\Rightarrow {e}_{H}^{2}=51$$⇒eH2​=51