The equation of the tangent of the ellipse 4x | vedclass

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(D) The given equation of the ellipse is $4x^2 + 9y^2 = 36$. Dividing by $36$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.Here,$a^2 = 9$ and $b^2 = 4$,

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$\sqrt{5}x - 3y + 9 = 0$

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(D) The given equation of the ellipse is $4x^2 + 9y^2 = 36$. Dividing by $36$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.Here,$a^2 = 9$ and $b^2 = 4$,so $a = 3$ and $b = 2$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.The coordinates of the latus rectum endpoints are $(\pm ae, \pm \frac{b^2}{a})$.For the second quadrant,$x = -ae = -3 	imes \frac{\sqrt{5}}{3} = -\sqrt{5}$ and $y = \frac{b^2}{a} = \frac{4}{3}$.The point is $P(-\sqrt{5}, \frac{4}{3})$.The equation of the tangent at $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.Substituting the values: $\frac{x(-\sqrt{5})}{9} + \frac{y(4/3)}{4} = 1$.$\Rightarrow -\frac{\sqrt{5}x}{9} + \frac{y}{3} = 1$.Multiplying by $-9$: $\sqrt{5}x - 3y = -9$,which simplifies to $\sqrt{5}x - 3y + 9 = 0$.

The equation of the tangent of the ellipse $4x^2 + 9y^2 = 36$ at the end of the latus rectum lying in the second quadrant is:

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