1
AP EAPCET 2025 - 26th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $\theta$ is the angle subtended by a latus rectum at the centre of the hyperbola having eccentricity $\frac{2}{\sqrt{7}-\sqrt{3}}$, then $\sin \theta=$

A

$\frac{1}{2} \tan \frac{\theta}{2}$

B

$2 \cos \frac{\theta}{2}$

C

$\frac{1}{\sin \frac{\theta}{2}+\cos \frac{\theta}{2}}$

D

$1-\cos \frac{\theta}{2}$

2
AP EAPCET 2025 - 26th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The tangent drawn at an extremity (in the first quadrant) of latus rectum of the hyperbola $\frac{x^2}{4}-\frac{y^2}{5}=1$ meets the $X$-axis and $Y$-axis at $A$ and $B$ respectively. If $O$ is the origin, then $(O A)^2-(O B)^2=$

A

$-\frac{20}{9}$

B

$\frac{16}{9}$

C

$-\frac{4}{9}$

D

$-\frac{4}{3}$

3
AP EAPCET 2025 - 27th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ passing through the point $(4,6)$ is 2 , then the equation of the tangent to this hyperbola at $(4,6)$ is

A

$2 x-3 y+10=0$

B

$3 x-2 y=0$

C

$x-2 y+8=0$

D

$2 x-y-2=0$

4
AP EAPCET 2025 - 27th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

A hyperbola passes through the point $P(\sqrt{2}, \sqrt{3})$ and has foci at $( \pm 2,0)$. Then, the point that lies on the tangent drawn to this hyperbola at $P$ is

A

$(\sqrt{3}, \sqrt{2})$

B

$(-\sqrt{2},-\sqrt{3})$

C

$(2 \sqrt{2}, 3 \sqrt{3})$

D

$(3 \sqrt{2}, 2 \sqrt{3})$

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