1
TS EAMCET 2022 (Online) 19th July Morning Shift
MCQ (Single Correct Answer)
+1
-0

The cartesian eql tion of the parabola $x=-2+2 t^2, y=2+4 t$ is

A

$y^2-8 x-4 y+12=0$

B

$y^2-8 x-4 y-12=0$

C

$y^2+8 x-4 y-12=0$

D

$y^2-8 x+4 y-12=0$

2
TS EAMCET 2022 (Online) 18th July Evening Shift
MCQ (Single Correct Answer)
+1
-0

The vertex and the focus of the parabola $2 x^2+5 y-6 x+1=0$ respectively, are

A

$\left(\frac{-3}{2}, \frac{7}{10}\right),\left(\frac{-3}{2}, \frac{53}{40}\right)$

B

$\left(\frac{-3}{2}, \frac{7}{10}\right),\left(\frac{-3}{2}, \frac{3}{40}\right)$

C

$\left(\frac{3}{2}, \frac{7}{10}\right),\left(\frac{3}{2}, \frac{53}{40}\right)$

D

$\left(\frac{3}{2}, \frac{7}{10}\right),\left(\frac{3}{2}, \frac{3}{40}\right)$

3
TS EAMCET 2022 (Online) 18th July Evening Shift
MCQ (Single Correct Answer)
+1
-0

The axis of a parabola is along the line $y=x$ and the distance of its vertex $A$ from $(0,0)$ is $\sqrt{2}$ and that of its focus $S$ from $(0,0)$ is $2 \sqrt{2}$. If $A$ and $S$ lie in first quadrant, then the equation of the parabola in parametric form is

A

$x=(t+1)^2, y=(t-1)^2$

B

$x=t^2, y=2 t$

C

$x=(t-\sqrt{2})^2, y=(t+\sqrt{2})^2$

D

$x=t^2+5, y=t^2-5$

4
TS EAMCET 2022 (Online) 18th July Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $y^2=16 x$ is the given parabola, then the point of intersection of the focal chord through the point $(2,2)$ and the double ordinate of length 24 is

A

$(3,1)$

B

$(9,-5)$

C

$(9,3)$

D

$(8,-4)$

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