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

If $A=(0,1), B=(1,2), C=(-2,1)$, then the equation of the locus of a point $P$ such that area of $\triangle P A B=$ area of $\triangle P A C$ is

A

$x^2-2 x y-3 y^2+2 x+6 y-3=0$

B

$x^2+2 x y-3 y^2+2 x+6 y-4=0$

C

$x^2-2 x y-3 y^2+2 x-6 y+4=0$

D

$x^2-2 x y+3 y^2-2 x+6 y-3=0$

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

If the latus rectum through one of the foci of a hyperbola $\frac{x^2}{9}-\frac{y^2}{b^2}=1$ subtends a right angle at the farther vertex of the hyperbola, then $b^2=$

A

4

B

16

C

25

D

27

3
TG EAPCET 2025 (Online) 3rd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Let $P, Q, R, S$ be the points of intersection of the circle $x^2+y^2=4$ and the hyperbola $x y=\sqrt{3}$. If $P=(\alpha, \beta)$ and $\alpha>\beta>0$, then the equation of the tangent drawn at $P$ to the hyperbola is

A

$x+y=2$

B

$x+\sqrt{3 y}=2 \sqrt{3}$

C

$\sqrt{3 x}+y=\sqrt{3}$

D

$x-y=0$

4
TG EAPCET 2025 (Online) 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If the tangent drawn at the point $P(3 \sqrt{2}, 4)$ on the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ meets its directrix at $Q(\alpha, \beta)$ in fourth quadrant, then $\beta=$

A

$\frac{5 \sqrt{2}-9}{4}$

B

$-\frac{9}{5}$

C

$\frac{12 \sqrt{2}-20}{5}$

D

$-\frac{5}{4}$

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