If 4x - 3y + k = 0 touches the ellipse 5x^{2} | vedclass

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(A) The given equation of the ellipse is $5x^{2} + 9y^{2} = 45$. Dividing by $45$,we get $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$. Here,$a^{2} = 9$ and

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$\pm 3\sqrt{21}$

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(A) The given equation of the ellipse is $5x^{2} + 9y^{2} = 45$. Dividing by $45$,we get $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$. Here,$a^{2} = 9$ and $b^{2} = 5$. The line $4x - 3y + k = 0$ can be written as $y = \frac{4}{3}x + \frac{k}{3}$. Comparing this with the line $y = mx + c$,we have $m = \frac{4}{3}$ and $c = \frac{k}{3}$. The condition for the line $y = mx + c$ to touch the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $c^{2} = a^{2}m^{2} + b^{2}$. Substituting the values,we get $(\frac{k}{3})^{2} = 9(\frac{4}{3})^{2} + 5$. $\frac{k^{2}}{9} = 9(\frac{16}{9}) + 5 = 16 + 5 = 21$. $k^{2} = 9 	imes 21 = 189$. $k = \pm \sqrt{189} = \pm 3\sqrt{21}$.

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

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