Question

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

• A. $\pm3\sqrt{21}$
• 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 = 2 1$
$ \Rightarrow k^{2}= 189$
$\Rightarrow k=\pm \sqrt{9\times21} = \pm3\sqrt{21}$