Question

If $4x-3y+k=0$ touches the ellipse $5x^2+9y^2=45,$ then $k=$

• A. $\pm 3\sqrt{21}$
• B. $3\sqrt{21}$
• C. $-3\sqrt{21}$
• D. $2\sqrt{21}$

Answer 1

Answer: (A)

$y=mx+c$ touches $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
''$c^2=a^2{m}^2+b^2$''
Given line $4x-3y+k=0$
$\Rightarrow y=\frac{4}{3}x+\frac{k}{3}$
Here $m=\frac{4}{3},C=\frac{k}{3}$
Ellipse is $\frac{x^2}{9}+\frac{y^2}{5}=1$
$\therefore a^2=9,b^2=5$
$\therefore \frac{k^2}{9}=9\left(\frac{16}{5}\right)+5=16+5=21$
$\Rightarrow k^2=189$
$\Rightarrow k=\pm \sqrt{9\times 21}=\pm 3\sqrt{21}$