MHTCET 22 Apr 2025 Shift 2 Paper

Concept:The foci of the given ellipse are first determined, and then these same foci are used to find the equation of the hyperbola with the given eccentricity.Explanation:For the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$, compare with the standard form.Here, $a^2 = 25$ and $b^2 = 9$, so $a = 5$ and $b = 3$.The eccentricity of the ellipse is:$$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$$Therefore, the foci of the ellipse are at:$$(\pm ae, 0) = \left(\pm 5 \cdot \frac{4}{5}, 0\right) = (\pm 4, 0)$$The foci of the hyperbola coincide with these, so its foci are also $(\pm 4, 0)$.For the hyperbola, the eccentricity is $e' = 2$, and the foci are $(\pm a'e', 0)$.Thus, $a'e' = 4$, which gives:$$a' = \frac{4}{e'} = \frac{4}{2} = 2$$Now, using the relation for a hyperbola:$$b'^2 = a'^2(e'^2 - 1) = 2^2(2^2 - 1) = 4 \times 3 = 12$$So the equation of the required hyperbola is:$$\frac{x^2}{a'^2} - \frac{y^2}{b'^2} = 1 \Rightarrow \frac{x^2}{4} - \frac{y^2}{12} = 1$$Answer:Option B: $\frac{x^2}{4} - \frac{y^2}{12} = 1$