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

Assertion (A) The length of the latus rectum of an ellipse is 4 . The focus and its corresponding directrix are respectively $(1,-2)$ and $3 x+4 y-15=0$. Then, its eccentricity is $\frac{1}{2}$.

Reason $(\mathrm{R})$ Length of the perpendicular drawn from focus of an ellipse to its corresponding directrix is $\frac{a\left(1-e^2\right)}{e}$.

Then, which one of the following is correct?

A

(A) and (R) are true and (R) is the correct explanation of (A)

B

(A) and (R) are true and (R) is not the correct explanation of (A)

C

(A) is true, (R) is false

D

(A) is false, (R) is true

2
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

$2 x-y-2=0$

3
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

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

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

The circumradius of the triangle formed by the points $(2,-1,1),(1,-3,-5)$ and $(3,-4,-4)$ is

A

$\frac{\sqrt{35}}{2}$

B

$\frac{\sqrt{25}}{3}$

C

$\sqrt{41}$

D

$\frac{\sqrt{41}}{2}$