Question

Tangents are drawn from $A(\alpha ,0)$ to the parabola $y^2=4x$ and the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$. Let the points of tangency on the ellipse be $P$ and $Q$ and on the parabola be $R$ and $S$. If $△APQ$ and $△ARS$ are equilateral, then which of the following is (are) CORRECT?

• A. $\alpha =-3$
• B. $\frac{\text{Area}\left(△APQ\right)}{\text{Area}\left(△ARS\right)}=\frac{b^4}{36}$
• C. $a\in (\frac{3}{2},3)$
• D. Distance between both of the chord of contact is $6-b^2$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\alpha =-3$
B $\frac{\text{Area}\left(△APQ\right)}{\text{Area}\left(△ARS\right)}=\frac{b^4}{36}$
C $a\in (\frac{3}{2},3)$
D Distance between both of the chord of contact is $6-b^2$

Equation of tangent in slope form on parabola,
$y=mx+\frac{1}{m}$
Putting $y=0$
$x=-\frac{1}{m^2}=\alpha \cdots \left(1\right)$
On ellipse,
$y=mx\pm \sqrt{a{m}^2+b^2}$
As they represent same tangents,
$\frac{1}{m^2}=a{m}^2+b^2\cdots \left(2\right)$
We know that $△APQ$ and $△ARS$ are equilateral.
So, $|m|=\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$
$\Rightarrow m^2=\frac{1}{3}$
From equation $\left(1\right)$,
$\alpha =-3$

So, from equation $\left(2\right)$,
$\frac{a^2}{3}+b^2=3\Rightarrow b^2=3-\frac{a^2}{3}\cdots \left(3\right)$
We know that, $a^2>b^2>0$
Using equation $\left(3\right)$,
$a^2>3-\frac{a^2}{3}>0\Rightarrow a^2>3-\frac{a^2}{3}3-\frac{a^2}{3}>0\Rightarrow 4a^2>99>a^2\Rightarrow a\in (\frac{3}{2},3)(\because a>0)$

Equation of chord of contact is
$T=0$
On the parabola,
$y\left(0\right)=2(x-3)\Rightarrow x=3$

On the ellipse,
$\frac{x(-3)}{a^2}+0=1\Rightarrow x=-\frac{a^2}{3}$
Distance between the chord of contacts
$=|3-(-\frac{a^2}{3})|=3+\frac{a^2}{3}$
$=3+3-b^2\left[\text{From (3)}\right]$
$=6-b^2$
Ratio of area of equilateral triangle $=$ Ratio of the square of their heights
$\frac{\text{Area}\left(△APQ\right)}{\text{Area}\left(△ARS\right)}=\frac{{(-3+\frac{a^2}{3})}^2}{(-3-3{)}^2}$
$=\frac{{(-3+3-b^2)}^2}{(-3-3{)}^2}=\frac{b^4}{36}\left[\text{From (3)}\right]$