Question

If one of the foci of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ coincide with the focus of the parabola $y^2=8x$ and they intersect at a point where the ordinate is double the abscissa, then the value of $\left[b^2\right]$ is
(where [.] represents greatest integer function)

• A. $5$
• B. $19$
• C. $10$
• D. $6$

Answer 1

The correct option is B $19$

Given parabola is $y^2=8x$
Focus $=(2,0)$
Given ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$
The focus coincide, so
$ae=2\Rightarrow a^2{e}^2=4\Rightarrow a^2-b^2=4\cdots \left(1\right)$
Let the point of intersection be $(t,2t)$
Now, putting this point on the parabola, we get
$4t^2=8t\Rightarrow t=0,2$
Rejecting $(0,0)$
Putting $(2,4)$ in the equation of the ellipse, we get
$\frac{x^2}{4+b^2}+\frac{y^2}{b^2}=1\text{from}\left(1\right)\Rightarrow \frac{4}{4+b^2}+\frac{16}{b^2}=1\Rightarrow b^4-16b^2-64=0\Rightarrow (b^2-8{)}^2=128\Rightarrow b^2-8=\pm 8\sqrt{2}\Rightarrow b^2=8+8\sqrt{2}(\because b^2>0)\therefore [b^2]=19$