Question

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

Answer 1

Answer: 2.00

$c=\pm \sqrt{a^2{m}^2+b^2}$
$c=\pm \sqrt{4(16)+1}=\pm \sqrt{65}$ (two values)