The values that $m$ can take, so that the straight line $y=4 x+m$ touches the curve $x^2+4 y^2=4$ is
The values that $m$ can take, so that the straight line $y=4 x+m$ touches the curve $x^2+4 y^2=4$ is
$\pm \sqrt{45}$
$\pm \sqrt{60}$
$\pm \sqrt{65}$
$\pm \sqrt{72}$
Solution
As the straight line, $y=m x+4$ touches the circle $x^2+4 y^2=4$
Put $y=4 x+m$ in $x^3+4 y^2=4$, we get
$\begin{aligned}
& x^2+4(4 x+m)^2=4 \\
& \Rightarrow \quad x^2+4\left(16 x^2+8 m x+m^2\right)=4 \\
& \Rightarrow \quad x^2+64 x^2+32 m x+4\left(m^2-1\right)=0 \\
& \Rightarrow \quad 65 x^2+32 m x+4\left(m^2-1\right)=0
\end{aligned}$
Since, line is tangent to the given curve
$\begin{array}{ll}
\Rightarrow & D=0 \\
\therefore & (32 m)^2-4(65)\left[4\left(m^2-1\right)\right]=0 \\
\Rightarrow & m= \pm \sqrt{65}
\end{array}$