Question

Given ellipse $x^2+4y^2=16$ and parabola $y^2-4x-4=0.$
The quadratic equation whose roots are the slopes of the common tangents to the parabola and the ellipse, is

• A. $3x^2-1=0$
• B. $5x^2-1=0$
• C. $15x^2+2x-1=0$
• D. $2x^2-1=0$

Answer 1

The correct option is B $5x^2-1=0$

Ellipse : $\frac{x^2}{16}+\frac{y^2}{4}=1$
Parabola : $y^2=4(x+1)$
Put $x+1=X$ and $y=Y$ in the parabola, so the tangent is $Y=mX+\frac{1}{m}=m(x+1)+\frac{1}{m}$
$y=mx+(m+\frac{1}{m})\cdots \left(1\right)$
If eqn$\left(1\right)$ is tangent to $\frac{x^2}{16}+\frac{y^2}{4}=1$ also, then $c^2=a^2{m}^2+b^2$
$\Rightarrow {(m+\frac{1}{m})}^2=16m^2+4$
$\Rightarrow m^2+\frac{1}{m^2}+2=16m^2+4$
$\Rightarrow 15m^2-\frac{1}{m^2}+2=0$
$\Rightarrow 15m^4+2m^2-1=0$
$\Rightarrow 15m^4+5m^2-3m^2-1=0$
$\Rightarrow 5m^2(3m^2+1)-(3m^2+1)=0$
$\Rightarrow (5m^2-1)(3m^2+1)=0$
$\Rightarrow 5m^2-1=0$