TS EAMCET 2020 (Online) 11th September Evening Shift | Ellipse Question 62 | Mathematics

If $x+2 y+k=0, k>0$ is a tangent to the ellipse $2 x^2+y^2=2$, then the equation of the normal to the given ellipse at $\left(\frac{1}{\sqrt{2}}, \frac{k}{3}\right)$, is

$\sqrt{2} x-2 y+1=0$

$3 \sqrt{2} x-y-2=0$

$2 \sqrt{2} x-5 y+3=0$

$\sqrt{2} x+3 y-4=0$

TS EAMCET 2020 (Online) 11th September Morning Shift

MCQ (Single Correct Answer)

-0

If $A=(1,2), B=(2,1)$ and $P$ is any point satisfying the condition $P A+P B=3$, then the equation of the locus of $P$ is

$16 x^2+7 y^2-64 x-48=0$

$x^2+10 x y+25 y^2-34 x-170 y=0$

$32 x^2+8 x y+32 y^2-108 x-108 y+99=0$

$4 x^2+12 x y+9 y^2-20 x-30 y=0$

TS EAMCET 2020 (Online) 11th September Morning Shift

MCQ (Single Correct Answer)

-0

If the sum of the distances from the foci to the centre $O(0,0)$ of an ellipse is $8 \sqrt{6}$ units and the area of the smallest rectangle in which that ellipse is inscribed is 80 sq. units, then the equation of such an ellipse is

$\frac{x^2}{100}+\frac{y^2}{64}=1$

$\frac{x^2}{100}+\frac{y^2}{16}=1$

$\frac{x^2}{10}+\frac{y^2}{4}=1$

$\frac{x^2}{100}+\frac{y^2}{4}=1$

TS EAMCET 2020 (Online) 11th September Morning Shift

MCQ (Single Correct Answer)

-0

The equation of the ellipse with directrix $3 x+4 y-5=0$, focus $(1,2)$ and eccentricity $1 / 2$, is

$x^2+84 y^2-24 x y-360 y+170 x+475=0$

$91 x^2+84 y^2-24 x y-170 x-360 y+475=0$

$91 x^2+84 y^2-24 x y-170 x+360 y+475=0$

$91 x^2+84 y^2-24 x y-170 x-360 y-475=0$

PREVIOUS NEXT

TS EAMCET Subjects

Browse all chapters by subject