1
MHT CET 2025 25th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The eccentricity of the hyperbola which passes through the points $(3,0)$ and $(3 \sqrt{2}, 2)$ is

A
$\sqrt{13}$
B
$\frac{\sqrt{13}}{4}$
C
$\frac{\sqrt{13}}{3}$
D
$\frac{\sqrt{13}}{2}$
2
MHT CET 2025 23rd April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If the tangent at the point $(2 \sec \theta, 3 \tan \theta)$ to the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$ is parallel to $3 x-y+4=0$, then the value of $\theta$ is

A
$45^{\circ}$
B
$60^{\circ}$
C
$30^{\circ}$
D
$90^{\circ}$
3
MHT CET 2025 22nd April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The foci of a hyperbola coincide with the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$. The equation of the hyperbola with eccentricity 2 is

A
$\frac{x^2}{12}-\frac{y^2}{4}=1$
B
$\frac{x^2}{4}-\frac{y^2}{12}=1$
C
$\frac{x^2}{12}-\frac{y^2}{16}=1$
D
$\frac{x^2}{16}-\frac{y^2}{12}=1$
4
MHT CET 2025 22nd April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The foci of a hyperbola coincide with the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$. The equation of the hyperbola with eccentricity 2 is

A
$\frac{x^2}{12}-\frac{y^2}{4}=1$
B
$\frac{x^2}{4}-\frac{y^2}{12}=1$
C
$\frac{x^2}{12}-\frac{y^2}{16}=1$
D
$\frac{x^2}{16}-\frac{y^2}{12}=1$
MHT CET Subjects
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