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

The slope of the non-vertical tangent drawn from the point $(3,4)$ to the circle $x^2+y^2=9$ is

A

$\frac{2}{3}$

B

$\frac{3}{2}$

C

$\frac{7}{24}$

D

$\frac{24}{7}$

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

If the acute angle between the circles $S \equiv x^2+y^2+2 k x+4 y-3=0$ and $S^{\prime} \equiv x^2+y^2-4 x+2 k y+9=0$ is $\cos ^{-1}\left(\frac{3}{8}\right)$ and the centre of $S^{\prime}=0$ lies in the first quadrant, then the radical axis of $S=0$ and $S^{\prime}=0$ is

A

$x-5 y+6=0$

B

$x-5 y-4=0$

C

$5 x-y-6=0$

D

$5 x-y-4=0$

3
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

$(3, \sqrt{2})$

4
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

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