The product of the slopes of the common tangents of the& ellipse x^...

Given, ellipse: $\dfrac{x2}{16} + \dfrac{y2}{4} = 1$ and& parabola: $y^{2} = 4 \left(x + 1\right)<br />Putting$x + 1 = X and Y = y$, we get,</p><p>$Y^{2} = 4X$</p><p>So, the equation of the tangent is$Y = mX + $\dfrac{1}{m} = m \left(x + 1\right) + \dfrac{1}{m}<br />$Y = mx + \left(m + \dfrac{1}{m}\right)$\ldots \left(i\right)$
If \left(i\right)$is also a tangent to$\dfrac{x^{2}}{16} + \dfrac{y^{2}}{4} =1$,</p><p>then,$c^{2} = a^{2}m^{2} + b^{2}$<br />$⇒ ${ \left( m + \dfrac{1}{m}\right)}$^{2} = 16m^{2} + 4$<br />$⇒ m^{2} + $\dfrac{1}{m^{2}} +2 = 16m^{2} + 4$
$⇒ 15m^{2} - \dfrac{1}{m^{2}} +2 = 0$
Let, $m^{2} = t > 0$

$⇒ 15t^{2} + 2t - 1 = 0$

$⇒ t = \dfrac{1}{5}, \dfrac{- 1}{3}$⇒ m^{2} = $\dfrac{1}{5}$

Hence, $m =$\pm \sqrt{$\dfrac{1}{5}$}</p><p>& </p>