1
TG EAPCET 2025 (Online) 2nd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $S$ and $S^{\prime}$ are the foci of an ellipse $\frac{x^2}{169}+\frac{y^2}{144}=1$ and the point $B$ lying on positive $Y$-axis is one end of its minor axis, then the incentre of the $\triangle S B S^{\prime}$ is

A

$\left(0, \frac{10}{3}\right)$

B

$\left(\frac{13}{3}, \frac{10}{3}\right)$

C

$\left(\frac{10}{3}, \frac{13}{3}\right)$

D

$\left(0, \frac{13}{3}\right)$

2
TG EAPCET 2025 (Online) 2nd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

One of the foci of an ellipse is $(2,-3)$ and its corresponding directrix is $2 x+y=5$. If the eccentricity of the ellipse is $\frac{\sqrt{5}}{3}$, then the coordinates of the other focus are

A

$(18,5)$

B

$(4,-2)$

C

$(-2,-5)$

D

$(-4,-6)$

3
TG EAPCET 2025 (Online) 2nd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If the product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes is $\frac{36}{13}$ and its eccentricity is $\frac{\sqrt{13}}{3}$, then $a-b=$

A

4

B

3

C

2

D

1

4
TG EAPCET 2025 (Online) 2nd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $A(0,3,4), B(1,5,6), C(-2,0,-2)$ are the vertices of a $\triangle A B C$ and the bisector of angle $A$ meets the side $B C$ at $D$, then $A D=$

A

$\frac{\sqrt{21}}{5}$

B

$\frac{\sqrt{42}}{10}$

C

10

D

4

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