Question

Find the number of values of $c$ such that the straight line $y=4x+c$ touches the curve $\frac{x^2}{4}+y^2=1$ .

Answer 1

Answer: 2

Given line $y=4x+c\Rightarrow m=4.$
Given ellipse $\frac{x^2}{4}+y^2=1\Rightarrow a^2=4\&b^2=1.$
We know that tangent condition is
$c=\pm \sqrt{a^2{m}^2+b^2}$
$c=\pm \sqrt{4\left(16\right)+1}=\pm \sqrt{65}$ (two values)