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

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)