1
AP EAPCET 2025 - 21st May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $L$ is the normal drawn to the parabola $y^2=8 x$ at the point $t=\frac{1}{\sqrt{2}}$, then the foot of the perpendicular drawn from the focus of the parabola on to the normal $L$ is

A

$(3,2)$

B

$(5, \sqrt{2})$

C

$(0, \sqrt{2})$

D

$(3, \sqrt{2})$

2
AP EAPCET 2025 - 21st May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If the tangents are drawn to the ellipse $x^2+2 y^2=2$, then the locus of the mid-points of the intercepts made by the tangents between the coordinate axes is

A

$\frac{x^2}{4}+\frac{y^2}{2}=1$

B

$\frac{x^2}{2}+\frac{y^2}{4}=1$

C

$\frac{1}{4 x^2}+\frac{1}{2 y^2}=1$

D

$\frac{1}{2 x^2}+\frac{1}{4 y^2}=1$

3
AP EAPCET 2025 - 21st May Morning Shift
MCQ (Single Correct Answer)
+1
-0

One of the latus recta of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ subtends an angle $2 \tan ^{-1}\left(\frac{3}{2}\right)$ at the centre of the hyperbola. If $b^2=36$ and $e$ is the eccentricity of the given hyperbola, then $\sqrt{a^2+e^2}=$

A

4

B

$\sqrt{14}$

C

6

D

$\sqrt{21}$

4
AP EAPCET 2025 - 21st May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If the equation of the hyperbola having $(8,3),(0,3)$ as foci and $\frac{4}{3}$ as eccentricity is $\frac{(x-\alpha)^2}{p}-\frac{(y-\beta)^2}{q}=1$, then $p+q=$

A

$\beta^2$

B

$\alpha+\beta$

C

$\alpha^2$

D

$\alpha \beta$