If the line 3x + 4y + k = 0 touches the ellip | vedclass

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(D) The given equation of the ellipse is $9x^2 + 16y^2 = 144$. Dividing by $144$, we get $\frac{x^2}{16} + \frac{y^2}{9} = 1$. Comparing with $\frac{x

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$\pm 12\sqrt{2}$

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(D) The given equation of the ellipse is $9x^2 + 16y^2 = 144$. Dividing by $144$, we get $\frac{x^2}{16} + \frac{y^2}{9} = 1$. Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we have $a^2 = 16$ and $b^2 = 9$. The condition for the line $y = mx + c$ to touch the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$. Rewrite the line $3x + 4y + k = 0$ as $4y = -3x - k$, which gives $y = -\frac{3}{4}x - \frac{k}{4}$. Here, $m = -\frac{3}{4}$ and $c = -\frac{k}{4}$. Substituting these values into the condition: $(-\frac{k}{4})^2 = 16(-\frac{3}{4})^2 + 9$. $\frac{k^2}{16} = 16(\frac{9}{16}) + 9$. $\frac{k^2}{16} = 9 + 9 = 18$. $k^2 = 18 	imes 16 = 288$. $k = \pm \sqrt{288} = \pm 12\sqrt{2}$.

If the line $3x + 4y + k = 0$ touches the ellipse $9x^2 + 16y^2 = 144$, then the value of $k$ is:

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