Question

$P:y^2=8x,E:\frac{x^2}{4}+\frac{y^2}{15}=1$

Point of contact of a common tangent to $P$ and $E$ on the ellipse is

• A. $(\frac{1}{2},\frac{15}{4})$
• B. $(-\frac{1}{2},\frac{15}{4})$
• C. $(\frac{1}{2},-\frac{15}{4})$
• D. $(-\frac{1}{2},-\frac{15}{4})$

Answer 1

Answer: (B), (D)

$(-\frac{1}{2},\frac{15}{4})$
$(-\frac{1}{2},-\frac{15}{4})$

Equation of tangent to the parabola $y^2=8x$ is

$y=mx+\frac{a}{m}=mx+\frac{2}{m}...\left(1\right),$ (here $a=2)$

Given ellipse is, $\frac{x^2}{4}+\frac{y^2}{15}=1$

$\Rightarrow a^2=4,b^2=15$

Also line (1) is also tangent to the ellipse, so using condition of tangency, $c^2=a^2{m}^2+b^2$

$\Rightarrow \frac{4}{m^2}=4m^2+15$

$\Rightarrow m^2=\frac{1}{4},m^2=-4$ (not possible)

$\Rightarrow m=\pm \frac{1}{2}$

Thus equation of common tangent is, $x\pm 2y+8=0$

Solving this equation with the ellipse we can get points of contact which are

$(-\frac{1}{2},\frac{15}{4})$ or $(-\frac{1}{2},-\frac{15}{4})$