Question

From a point on the axis of $x$ common tangents are drawn to the parabola $y^2=4x$ and the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=(a>b>0).$ If these tangents form an equilateral triangle with their chord of contact w.r.t. parabola, then set of exhaustive values of $a$ is

• A. $(0,3)$
• B. $(\frac{3}{2},3)$
• C. $(1,\frac{3}{2})$
• D. $(0,\frac{3}{2})$

Answer 1

The correct option is B $(\frac{3}{2},3)$

$P$ is a point on axis of $x$ and $PA,PB$ are tangents drawn from $P$ to parabola $y^2=4x,a=1$ so that $△PAB$ is equilateral.
Any tangent to parabola $y=mx+\frac{1}{m}$ where $m=\tan (\pm {30}^{\circ })=\pm \frac{1}{\sqrt{3}}$
But these tangents are also tangents to the ellipse.
Hence, condition of tangency $c^2=a^2{m}^2+b^2$ gives $\frac{1}{m^2}=a^2{m}^2+b^2$ or $3=\frac{a^2}{3}+b^2$ or $9=a^2+3a^2(1-e^2)$

$\therefore \frac{9-a^2}{3a^2}=1-e^2$

$e^2=1-\frac{9-a^2}{3a^2}$

$e^2=\frac{4a^2-9}{3a^2}$

Since $0<e^2<1$

$0<\frac{4a^2-9}{3a^2}<1$
$0<4a^2-9<3a^2$

$a^2<9$ or $\frac{9}{4}<a^2<9$

$\therefore \frac{3}{2}<a<3$ $\because a^2<x^2<b^2\Rightarrow a<x<b$ $-b<x<-a$
Ans: B