Let the foci and length of the latus rectum of an ellipse \fracx^2a^2+\fracy^2b^2=1, a>b b e( \pm 5,0) and \sqrt50 , respectively. Then, the square of the eccentricity of the hyperbola \fracx^2b^2-\fracy^2a^2 b^2=1 equals

Hyperbola PYQ — JEE Main 2024 · 31 Jan · Shift 1 · Q57 with Solution

View written solutionFree

Correct answer: 51

The ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad a>b$ with foci at $(\pm 5,0).$

So its focal distance is $c=5,$ and for an ellipse, $c^2=a^2-b^2.$ Hence, $a^2-b^2=25. \tag{1}$

For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ the length of the latus rectum is $\frac{2b^2}{a}.$

Given this equals $\sqrt{50}=5\sqrt{2}$ , so $\frac{2b^2}{a}=5\sqrt{2}. \tag{2}$

From (2), $b^2=\frac{5a\sqrt{2}}{2}. \tag{3}$

Substitute (3) into (1): $a^2-\frac{5a\sqrt{2}}{2}=25.$ Multiply by 2: $2a^2-5\sqrt{2}a-50=0.$

Solve this quadratic in $a$ :

=\frac{5\sqrt{2}\pm \sqrt{50+400}}{4} =\frac{5\sqrt{2}\pm 15\sqrt{2}}{4}.$$ Since $a>0$, $$a=\frac{20\sqrt{2}}{4}=5\sqrt{2}.$$ Then from (3), $$b^2=\frac{5(5\sqrt{2})\sqrt{2}}{2}=\frac{25\cdot 2}{2}=25.$$ Also, $$a^2=(5\sqrt{2})^2=50.$$ 4. **Interpret the hyperbola** The given hyperbola is $$\frac{x^2}{b^2}-\frac{y^2}{a^2b^2}=1.$$ Compare with standard form $$\frac{x^2}{A^2}-\frac{y^2}{B^2}=1.$$ So, $$A^2=b^2=25, \qquad B^2=a^2b^2=50\cdot 25=1250.$$ For a hyperbola, $$e^2=1+\frac{B^2}{A^2}.$$ Thus, $$e^2=1+\frac{1250}{25}=1+50=51.$$ 5. **Final answer** Therefore, the square of the eccentricity is $$\boxed{51}.$$