1
TS EAMCET 2022 (Online) 18th July Morning Shift
MCQ (Single Correct Answer)
+1
-0

The parametric equations of the ellipse whose focii are $(-3,0),(9,0)$ and eccentricity is $\frac{1}{3}$, are

A

$x=3+12 \sqrt{2} \cos \theta, y=18 \sin \theta$

B

$x=3+18 \cos \theta, y=12 \sqrt{2} \sin \theta$

C

$x=18 \cos \theta, y=3+12 \sqrt{2} \sin \theta$

D

$x=3+4 \sqrt{2} \cos \theta, y=18 \sin \theta$

2
TS EAMCET 2020 (Online) 14th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $a \alpha^2+b \beta^2+c \alpha \beta+d=0$ is the transformed equation of $4 x^2+\sqrt{3} x y+5 y^2-4=0$ obtained by using $\alpha=\frac{\sqrt{3}}{2} x+\frac{y}{2}$ and $\beta=-\frac{x}{2}+\frac{\sqrt{3}}{2} y$, then $c(a+b+d)=$

A

0

B

$13 \sqrt{3}$

C

$5 \sqrt{3}$

D

6

3
TS EAMCET 2020 (Online) 14th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

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

A

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

B

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

C

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

D

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

4
TS EAMCET 2020 (Online) 14th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

The area (in sq. units) of the quadrilateral formed by the tangents drawn at the end points of the latus rectum to the ellipse $S \equiv \frac{x^2}{16}+\frac{y^2}{12}=1$ is

A

96

B

16

C

128

D

64

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