Let the foci and length of the latus rectum of an ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$$ be $$(\pm 5, 0)$$ and $$\sqrt{50}$$, respectively. Then, the square of the eccentricity of the hyperbola $$\frac{x^2}{b^2} - \frac{y^2}{a^2 b^2} = 1$$ equals

For the ellipse $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1,\; a\gt b$$ the standard facts are

• Distance of each focus from the centre: $$c,\; c^2=a^2-b^2$$
• Length of the latus-rectum: $$L=\dfrac{2b^2}{a}$$

The question gives

$$c=5 \;\; \bigl(\text{foci } (\pm5,0)\bigr), \qquad L=\sqrt{50}$$

Step 1: Use the latus-rectum length.

$$\dfrac{2b^2}{a}= \sqrt{50}\;\; \Longrightarrow\;\; b^2=\dfrac{a\sqrt{50}}{2} \quad -(1)$$

Step 2: Use the focus relation.

$$a^2-b^2=25 \quad -(2)$$

Step 3: Substitute $$b^2$$ from $$(1)$$ into $$(2)$$.

$$a^2-\dfrac{a\sqrt{50}}{2}=25$$

Multiply by $$2$$:

$$2a^2-a\sqrt{50}-50=0$$

Step 4: Solve the quadratic in $$a$$.

Using the quadratic formula, $$a=\dfrac{\sqrt{50}\pm\sqrt{(\sqrt{50})^2-4(2)(-50)}}{4}$$

Discriminant: $$(\sqrt{50})^2-4(2)(-50)=50+400=450$$, so $$\sqrt{450}=3\sqrt{50}$$.

Hence $$a=\dfrac{\sqrt{50}\pm3\sqrt{50}}{4}$$. Since $$a\gt0$$, take the positive sign:

$$a=\sqrt{50}, \qquad\Longrightarrow\qquad a^2=50$$

Step 5: Eccentricity of the hyperbola.

The given hyperbola is $$\dfrac{x^2}{b^2}-\dfrac{y^2}{a^2b^2}=1$$

Compare with the standard form $$\dfrac{x^2}{A^2}-\dfrac{y^2}{B^2}=1$$, where

$$A^2=b^2, \qquad B^2=a^2b^2$$

For a hyperbola, $$e^2=1+\dfrac{B^2}{A^2}$$.

Therefore

$$e_{\text{hyper}}^{\,2}=1+\dfrac{a^2b^2}{b^2}=1+a^2$$

Using $$a^2=50$$, we get

$$e_{\text{hyper}}^{\,2}=1+50=51$$

Thus, the square of the eccentricity of the hyperbola equals $$51$$.