Question

Common tangents are drawn to the parabola $y^2=4x$ and the ellipse $3x^2+8y^2=48$ touching the parabola $A$ and $B$ and the ellipse at $C$ and $D$. Area of quadrilateral $ABCD$ (in sq.units) is:

• A. $55\sqrt{2}$
• B. $50$
• C. $50\sqrt{2}$
• D. $100\sqrt{2}$

Answer 1

The correct option is A $55\sqrt{2}$

Let $y=mx+\frac{1}{m}$ be tangent to parabola $y^2=4x$ it will touch the ellipse
$\frac{x^2}{4^2}+\frac{y^2}{(\sqrt{6}{)}^2}=1$
if $\frac{1}{m^2}=16m^2+6$
[using $c^2=a^2{m}^2+b^2$]
$\Rightarrow (8m^2-1)(2m^2+1)=0$
$\Rightarrow 16m^4+6m^2-1=0$
$\Rightarrow m=\pm \frac{1}{2\sqrt{2}}$
We know that a tangent of slope $m$ touches the parabola $y^2=4ax$ at $(\frac{a}{m^2},\frac{2a}{m}).$
Coordinates of the points of contact of common tangents to the parabola are
$\therefore A(8,4\sqrt{2})$ and $B(8,-4\sqrt{2})$
We also know that a tangent of slope $m$ touches the ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at
$(\mp \frac{a^{2}m}{\sqrt{a^2{m}^2+b^2}},\pm \frac{b^2}{\sqrt{a^2{m}^2+b^2}})$
Coordinates of the points of contact of common tangents to the ellipse are
$C(-2,\frac{3}{\sqrt{2}})$ and $D(-2,\frac{-3}{\sqrt{2}})$
Clearly thr quadrilateral $ABCD$ is a tarpezium we have $AB=8\sqrt{2},CD=\frac{6}{\sqrt{2}}$
and the distance between $AB$ and $CD$ is
$PQ=8+2=10$
$\therefore$ Area of quadrilateral $ABCD$
$\frac{1}{2}(AB+CD)\times PQ=55\sqrt{2}$ Sq. units