Question

The tangents are drawn to $\frac{x^2}{28}+\frac{y^2}{16}=1$ making an angle ${60}^{\circ }$ with positive $x-$axis, then the distance between tangents is units

Answer 1

For the given ellipse $\frac{x^2}{28}+\frac{y^2}{16}=1$
$a^2=28,b^2=16$

Equation of tangent in slope form is
$y=mx\pm \sqrt{a^2{m}^2+b^2}\cdots \left(1\right)\theta ={60}^{\circ }\Rightarrow m=\tan \theta =\sqrt{3}$
Putting values in equation $\left(1\right)$, we get
$y=\sqrt{3}x\pm \sqrt{3\times 28+16}\therefore y=\sqrt{3}x\pm 10$

Hence, the required distance is
$d=\frac{|10+10|}{\sqrt{1+3}}=10\text{units}$