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

Conic Sections

Solution:

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