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The values of c such that the line $y=4 x+c$ touches the ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$ is
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Verified Answer
The correct answer is:
$\pm \sqrt{65}$
$y=4 x+c$
$\Rightarrow \quad \frac{x^2}{4}+\frac{y^2}{1}=1$
If $y=m x+c$ is tangent then
$\begin{aligned} & c^2=a^2 m^2+b^2 \\ & \Rightarrow \quad c^2=4(4)^2+1^2=65 \\ & \therefore \quad c^2= \pm \sqrt{65} .\end{aligned}$
$\Rightarrow \quad \frac{x^2}{4}+\frac{y^2}{1}=1$
If $y=m x+c$ is tangent then
$\begin{aligned} & c^2=a^2 m^2+b^2 \\ & \Rightarrow \quad c^2=4(4)^2+1^2=65 \\ & \therefore \quad c^2= \pm \sqrt{65} .\end{aligned}$
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