Question

Two lines are drawn from point $P(\alpha ,\beta )$ which touches $y^2=8x$ at $A,B$ and touches $\frac{x^2}{4}+\frac{y^2}{6}=1$ at $C,D$, then

• A. $\alpha +\beta =-4$
• B. $\alpha \beta =4$
• C. Area of $△PAB=128\sqrt{2}$
• D. Area of $△PAB=32\sqrt{2}$

Answer 1

Answer: (A), (D)

$\frac{x^2}{4}+\frac{y^2}{6}=1$ and $y^2=8x$
Tangent to the ellipse
$y=mx\pm \sqrt{4m^2+6}$
Tangent to the parabola
$\text{ Tangent to the parabola }$
$y=mx+\frac{2}{m}$
$\therefore \frac{2}{m}=\pm \sqrt{4m^2+6}$
$4=m^2\left(4m^2+6\right)$
$2m^4+3m^2-2=0$
$\left(m^2+2\right)\left(2m^2-1\right)=0\Rightarrow m=\pm \frac{1}{\sqrt{2}}$
$\therefore y=\frac{x}{\sqrt{2}}+2\sqrt{2}$
$\sqrt{2}y=x+4\Rightarrow \alpha =-4,\beta =0$
$\mathrm{Ar}(\Delta PAB)=\frac{s_1^{3/2}}{2|a|}=\frac{(0+32{)}^{3/2}}{2\cdot 2}=\frac{32\cdot 4\sqrt{2}}{4}=32\sqrt{2}$