\[\begin{array}{1 1}(1)\pm 2\sqrt{3}&(2)\pm 6\\(3)36&(4)\pm4\end{array}\]

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

The equation of the ellipse is

$4x^2+8y^2=32$

$\large\frac{4x^2}{32} + \frac{8y^2}{32} $$=1$

$\large\frac{x^2}{8} + \frac{y^2}{4} $$=1$-----(1)

The equation of the line is

$2x-y+c=0$

$y=2x+c$ ------(2)

The condition for the line $y=mx+c$ to touch the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{^2}$$=1$ is $c^2=a^2m^2+^2$

From (1) and (2)

$a^2=8, b^2=4,m=2,c=c$

$c^2=8 \times 2^2+4$

$c^2=32+4=36$

$c= \pm 6$

Hence 2 is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...