Given,
Ellipse a2x2+b2y2=1
And its focii ≡(±5,0) and its latusrectum a2b2=50
So, ae=5 and b2=252a
Now, using the formula eccentricity we get,
b2=a2(1−e2)=252a
⇒a(1−e2)=252 asa=0
⇒e5(1−e2)=252
⇒2−2e2=e
⇒2e2+e−2=0
⇒2e2+2e−e−2=0
⇒2e(e+2)−1(1+2)=0
⇒(e+2)(2e−1)=0
∴e=−2;e=21
⇒a=52 and b=5
Now, finding eccentricity of hyperbola,
b2x2−a2b2y2=1
We know that formula of eccentricity of hyperbola is given by, B2=A2(eH2−1) where {A}^{2}={b}^{2}&{B}^{2}={a}^{2}{b}^{2}
⇒a2b2=b2(eH2−1)
⇒a2=(eH2−1)
⇒50=(eH2−1)
⇒eH2=51