1
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

2
TG EAPCET 2024 (Online) 11th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
$P(\theta)$ is a point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1, S$ is its $\mathrm{fOO}_{4 /}$ lying on the positive $X$-axis and $Q=(0,1)$. If $S Q=\sqrt{26}$ and $S P=6$, then $\theta=$
A
$\frac{\pi}{6}$
B
$\frac{\pi}{4}$
C
$\frac{\pi}{3}$
D
$\cos ^{-1}\left(\frac{?}{3}\right)$
3
TG EAPCET 2024 (Online) 10th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
If the tangent drawn at a point $P(t)$ on the hyperbola $x^{2}-y^{2}=c^{2}$ cuts $X$-axis at $T$ and the normal drawn at the same point $P$ cuts the $Y$-axis at $N$, then the equation of the locus of the mid-point of $T N$ is
A
$\frac{c^{2}}{4 x^{2}}-\frac{y^{2}}{c^{2}}=1$
B
$\frac{x^{2}}{c^{2}}-\frac{y^{2}}{4 c^{2}}=1$
C
$\frac{x^{2}}{4 c^{2}}+\frac{y^{2}}{c^{2}}=1$
D
$x^{2}+y^{2}=4 c^{2}$
4
TG EAPCET 2024 (Online) 10th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
The slope of the tangent drawn from the point $(1,1)$ to the hyperbola $2 x^2-y^2=4$ is
A
2
B
$\frac{-2 \pm \sqrt{6}}{2}$
C
$-1 \pm \sqrt{6}$
D
$\frac{-2 \pm \sqrt{3}}{2}$

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