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If the line $y=2 x+c$ be a tangent to the ellipse $\frac{x^2}{8}+\frac{y^2}{4}=1$, then $c=$
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The correct answer is:
$\pm 6$
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad y=m \underline{x+c}$
Condition for tangency :- $c^2=a^2 m^2+b^2$
$\frac{x^2}{8}+\frac{y^2}{4}=1$
$\begin{aligned} & a^2=8 \\ & b^2=4\end{aligned}$
$y=2 x+c$
$m=-2$
$c^2=8(2)^2+4 \quad \Rightarrow \quad c^2=8 \times 4+4$
$\begin{aligned} & c^2=8(2)^2+4 \quad \Rightarrow c^2=8 \times 4+4 \\ & c^2=32+4 \\ & c^2=36 \\ & c= \pm \sqrt{36} \\ & c= \pm 6\end{aligned}$
Condition for tangency :- $c^2=a^2 m^2+b^2$
$\frac{x^2}{8}+\frac{y^2}{4}=1$
$\begin{aligned} & a^2=8 \\ & b^2=4\end{aligned}$
$y=2 x+c$
$m=-2$
$c^2=8(2)^2+4 \quad \Rightarrow \quad c^2=8 \times 4+4$
$\begin{aligned} & c^2=8(2)^2+4 \quad \Rightarrow c^2=8 \times 4+4 \\ & c^2=32+4 \\ & c^2=36 \\ & c= \pm \sqrt{36} \\ & c= \pm 6\end{aligned}$
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