1
TG EAPCET 2024 (Online) 10th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
$S$ is the focus of the ellips $\frac{x^2}{25}+\frac{y^2}{b^2}=1,(b<5)$ lying on the negative $X$-axis and $P(\theta)$ is a point on this ellipes. If the distance between the foci of this ellipse is 8 and $S^{\prime} P=7$, then $\theta=$
A
$\frac{\pi}{6}$
B
$\frac{\pi}{3}$
C
$\frac{\pi}{4}$
D
$\frac{2 \pi}{3}$
2
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
$\frac{-2 \pm \sqrt{3}}{2}$
3
TG EAPCET 2024 (Online) 10th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
$A(2,3, k), B(-1, k,-1)$ and $C(4,-3,2)$ are the vertices of $\triangle A B C$. If $A B=A C$ and $k>0$, then $\triangle A B C$ is
A
an equilateral triangle
B
a right-angled isosceles triangle
C
an isosceles triangle but not right angled
D
an obtuse angled isosceles triangle
4
TG EAPCET 2024 (Online) 10th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
If $A(1,2,-3), B(2,3,-1)$ and $C(3,1,1)$ are the vertices of $\triangle A B C$, then $\left|\frac{-\cos A}{\cos B}\right|=$
A
$\frac{3 \sqrt{3}}{4 \sqrt{2}}$
B
$\frac{3 \sqrt{3}}{\sqrt{7}}$
C
$\frac{4 \sqrt{2}}{3 \sqrt{3}}$
D
$\frac{\sqrt{7}}{3 \sqrt{3}}$
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