Question

Consider two straight lines, each of which is tangent to both the circle $x^2+y^2=\frac{1}{2}$ and the parabola $y^2=4x$. Let these lines intersect at the point $Q$. Consider the ellipse whose center is at the origin $O(0,0)$ and whose semi-major axis is $OQ$. If the length of the minor axis of this ellipse is $\sqrt{2}$, then the which of the following statement(s) is (are) TRUE?

• A. For the ellipse, the eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $1$
• B. For the ellipse, the eccentricity is $\frac{1}{2}$ and the length of the latus rectum is $\frac{1}{2}$
• C. The area of the region bounded by the ellipse between the lines $x=\frac{1}{\sqrt{2}}$ and $x=1$ is $\frac{1}{4\sqrt{2}}(\pi -2)$
• D. The area of the region bounded by the ellipse between the lines $x=\frac{1}{\sqrt{2}}$ and $x=1$ is $\frac{1}{16\sqrt{2}}(\pi -2)$

Answer 1

Answer: (A), (C)

The area of the region bounded by the ellipse between the lines $x=\frac{1}{\sqrt{2}}$ and $x=1$ is $\frac{1}{4\sqrt{2}}(\pi -2)$
For the ellipse, the eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $1$

Let equation of common tangent is $y=mx+\frac{1}{m}$
$\therefore \left|\frac{0+0+\frac{1}{m}}{\sqrt{1+m^2}}\right|=\frac{1}{\sqrt{2}}\Rightarrow m^4+m^2-2=0\Rightarrow m=\pm 1$
Equation of common tangents are $y=x+1$ and $y=-x+1$
point $Q$ is $(-1,0)$
$\therefore$ Equation of ellipse is $\frac{x^2}{1}+\frac{y^2}{1/2}=1$
$\left(A\right)$ $e=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$ and $LR=\frac{2b^2}{a}=1$
$\left(C\right)$ Area $2.{\int }_{1/\sqrt{2}}^1\frac{1}{\sqrt{2}}.\sqrt{1-x^2}dx=\sqrt{2}{[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x]}_{1/\sqrt{2}}^1$
$=\sqrt{2}[\frac{\pi }{4}-(\frac{1}{4}+\frac{\pi }{8})]=\sqrt{2}(\frac{\pi }{8}-\frac{1}{4}]=\frac{\pi -2}{4\sqrt{2}}$
Correct answers are $\left(A\right)$ and $\left(C\right)$.