If the point of intersections of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the circle $x^2+y^2=4 b, b>4$ lie on the curve $y^2=3 x^2$, then $b$ is equal to :
Select the correct option:
A
12
B
5
C
6
D
10
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$y^2=3 x^2$ and $x^2+y^2=4 b$ Solve both we get
So $\mathrm{x}^2=\mathrm{b}$
$$
\frac{x^2}{16}+\frac{3 x^2}{b^2}=1, \frac{b}{16}+\frac{3}{b}=1
$$
$$
\begin{aligned}
& b^2-16 b+48=0 \\
& (b-12)(b-4)=0 \\
& b=12, b>4
\end{aligned}
$$
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation. Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇
