JEE Advanced 2018 Paper 2Question 1
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 which of the following statement(s) is (are) TRUE?
Ellipse
Solution
Concept. Find the common tangents, their meeting point $Q$, then the ellipse. The area comes from a standard integral. Formulas used. (1) Tangent to $y^2=4x$ with slope $m$: $y=mx+\frac1m$(2) Line touches a circle iff its distance from the centre equals the radius(3) $\int\sqrt{1-x^2}\,dx=\frac{x\sqrt{1-x^2}}2+\frac12\sin^{-1}x$Step 1 - Common tangents. By (1) and (2), distance of $mx-y+\frac1m=0$ from $O$ is $\frac1{\sqrt2}$: $$\frac{1/|m|}{\sqrt{1+m^2}}=\frac1{\sqrt2}\Rightarrow m^2(1+m^2)=2\Rightarrow(m^2+2)(m^2-1)=0$$ So $m=\pm1$: $y=x+1$ and $y=-x-1$. They meet at $Q(-1,0)$. Step 2 - Ellipse. $a=OQ=1$; minor axis $2b=\sqrt2\Rightarrow b^2=\frac12$. Ellipse: $x^2+2y^2=1$. Step 3 - A and B. $e^2=1-\frac{b^2}{a^2}=\frac12\Rightarrow e=\frac1{\sqrt2}$; latus rectum $\frac{2b^2}a=1$. A is true, B is false.Step 4 - Area. $y=\pm\sqrt{\frac{1-x^2}2}$, so $$\text{Area}=2\int_{1/\sqrt2}^{1}\sqrt{\tfrac{1-x^2}2}\,dx=\sqrt2\int_{1/\sqrt2}^{1}\sqrt{1-x^2}\,dx$$ By (3): value at $1$ is $\frac\pi4$; at $\frac1{\sqrt2}$ it is $\frac14+\frac\pi8$. Difference $=\frac\pi8-\frac14=\frac{\pi-2}8$. $$\text{Area}=\sqrt2\cdot\frac{\pi-2}8=\frac{\pi-2}{4\sqrt2}$$ C is true, D is false. Answer: A, C