JEE Advanced 2018 · Paper 2, Mathematics question 4

Common tangents. A tangent to $y^2=4x$ is $y=mx+\dfrac1m$. Its distance from the origin must be $\dfrac1{\sqrt2}$:
$$\frac{1/|m|}{\sqrt{1+m^2}}=\frac1{\sqrt2}\implies m^4+m^2-2=0\implies m^2=1.$$
The lines $y=x+1$ and $y=-x-1$ meet at $Q(-1,0)$.

Ellipse. $a=OQ=1$ and $2b=\sqrt2$, so $b=\dfrac1{\sqrt2}$.
$$e=\sqrt{1-\frac12}=\frac1{\sqrt2},\qquad\text{latus rectum}=\frac{2b^2}a=1.$$
(A) is true and (B) is false.

Area between $x=\frac1{\sqrt2}$ and $x=1$:
$$2b\int_{1/\sqrt2}^1\sqrt{1-x^2}\,dx=\sqrt2\cdot\frac12\left[x\sqrt{1-x^2}+\sin^{-1}x\right]_{1/\sqrt2}^1=\frac{\sqrt2}2\left(\frac\pi2-\frac12-\frac\pi4\right)=\frac{\pi-2}{4\sqrt2}.$$
(C) is true and (D) is false.

(A), (C)