If the point of intersections of the ellipse

Question:

If the point of intersections of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ and the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=4 \mathrm{~b}, \mathrm{~b}>4$

lie on the curve $y^{2}=3 x^{2}$, then $b$ is equal to:

  1. (1) 12

  2. (2) 5

  3. (3) 6

  4. (4) 10


Correct Option: 1

Solution:

$y^{2}=3 x^{2}$

and $x^{2}+y^{2}=4 b$

Solve both we get

so $x^{2}=b$

$\frac{x^{2}}{16}+\frac{3 x^{2}}{b^{2}}=1$

$\frac{b}{16}+\frac{3}{b}=1$

$b^{2}-16 b+48=0$

$(b-12)(b-4)=0$

$b=12, b>4$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now