TS EAMCET 2022 (Online) 19th July Morning Shift | Parabola Question 43 | Mathematics

The point to which the origin is to be shifted by translation of axes so that the transformed equation of $y^2+4 y+8 x-2=0$ will not contain $y$ term and constant term is

$\left(\frac{3}{4},-2\right)$

$\left(\frac{-3}{4},-2\right)$

$\left(2, \frac{3}{4}\right)$

$\left(-2, \frac{-3}{4}\right)$

TS EAMCET 2022 (Online) 19th July Morning Shift

MCQ (Single Correct Answer)

-0

Statement $14 x^2+y^2-4 x y-30 x-50 y+40=0$ is the equation of parabola having $(2,3)$ as its focus and $x+2 y+5=0$ as its directrix.

Statement II The equation of the directrix of the parabola $x^2-4 x+16 y+52=0$ is $y+1=0$

Which of the above statements is (are) true?

Statement I is true, but Statement II is false

Statement II is true, but Statement I is false

Both Statement I and Statement II are true

Both Statement I and Statement II are false

TS EAMCET 2022 (Online) 19th July Morning Shift

MCQ (Single Correct Answer)

-0

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

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

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

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

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

TS EAMCET 2022 (Online) 18th July Evening Shift

MCQ (Single Correct Answer)

-0

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

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

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

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

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

PREVIOUS NEXT

TS EAMCET Subjects

Browse all chapters by subject