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If $4 x-3 y+k=0$ touches the ellipse $5 x^{2}+9 y^{2}=45$, then $k$ is equal to
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Verified Answer
The correct answer is:
$\pm 3 \sqrt{21}$
If line $4 x-3 y+k=0$ touches the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$, then
$\frac{k}{3}=\sqrt{9 \times\left(\frac{4}{3}\right)^{2}+5}=\pm \sqrt{21}$
$\Rightarrow$
$k=\pm 3 \sqrt{21}$
$\frac{k}{3}=\sqrt{9 \times\left(\frac{4}{3}\right)^{2}+5}=\pm \sqrt{21}$
$\Rightarrow$
$k=\pm 3 \sqrt{21}$
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