MHTCET 22 Apr 2025 Shift 2 Paper

Concept:The foci of the ellipse give the value of $c$ for the hyperbola, and the eccentricity $e$ then gives $a$.Explanation:For the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, we have $a^{2}=25$ and $b^{2}=9$.Its eccentricity is $e_{e}=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$.Therefore, the foci of the ellipse are at $(\pm ae_{e},0)=(\pm 4,0)$.Since the hyperbola has the same foci, its foci are also $(\pm 4,0)$, so $c=4$.Given the eccentricity of the hyperbola is $e=2$, we use $c=ae$, giving $a=\frac{c}{e}=\frac{4}{2}=2$.For a hyperbola, $c^{2}=a^{2}+b^{2}$, so $16=4+b^{2}$, which gives $b^{2}=12$.Thus, the hyperbola has $a^{2}=4$ and $b^{2}=12$, with its transverse axis along the $x$-axis.Hence, its equation is $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$.Answer:$\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$Therefore, the correct option is B.