If 4x + 2y + n = 0 is a normal to the ellipse | vedclass

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(D) The equation of the line is $4x + 2y + n = 0$,which can be rewritten as $y = -2x - \frac{n}{2}$.Comparing this with $y = mx + c$,we get $m = -2$ a

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$\pm 8$

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(D) The equation of the line is $4x + 2y + n = 0$,which can be rewritten as $y = -2x - \frac{n}{2}$.Comparing this with $y = mx + c$,we get $m = -2$ and $c = -\frac{n}{2}$.The equation of the ellipse is $\frac{x^2}{36} + \frac{y^2}{16} = 1$,so $a^2 = 36$ and $b^2 = 16$.The condition for the line $y = mx + c$ to be a normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = \frac{(a^2 - b^2)^2 m^2}{a^2 + b^2 m^2}$.Substituting the values: $(-\frac{n}{2})^2 = \frac{(36 - 16)^2 (-2)^2}{36 + 16(-2)^2}$.$\frac{n^2}{4} = \frac{(20)^2 	imes 4}{36 + 16(4)} = \frac{400 	imes 4}{36 + 64} = \frac{1600}{100} = 16$.$n^2 = 16 	imes 4 = 64$.Therefore,$n = \pm 8$.

If $4x + 2y + n = 0$ is a normal to the ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$,then $n = $

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