The equation of the normal to the curve 4x^2 | vedclass

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(C) The given curve is $4x^2 + 9y^2 = 36$. Dividing by $36$, we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.This is an ellipse of the form $\frac{x^2}{a^2

Vedclass · Accepted Answer

$3\sqrt{2}x + 2\sqrt{2}y - 5 = 0$

Vedclass · Answer

(C) The given curve is $4x^2 + 9y^2 = 36$. Dividing by $36$, we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.This is an ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a^2 = 9$ $(a = 3)$ and $b^2 = 4$ $(b = 2)$.The parametric coordinates of any point on the ellipse are $(a \cos 	heta, b \sin 	heta) = (3 \cos 	heta, 2 \sin 	heta)$.The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $	heta$ is given by $ax \sec 	heta - by \operatorname{cosec} 	heta = a^2 - b^2$.Substituting $a = 3$, $b = 2$, and $	heta = \frac{7\pi}{4}$:$3x \sec(\frac{7\pi}{4}) - 2y \operatorname{cosec}(\frac{7\pi}{4}) = 9 - 4$Since $\sec(\frac{7\pi}{4}) = \sec(2\pi - \frac{\pi}{4}) = \sec(\frac{\pi}{4}) = \sqrt{2}$ and $\operatorname{cosec}(\frac{7\pi}{4}) = \operatorname{cosec}(2\pi - \frac{\pi}{4}) = -\operatorname{cosec}(\frac{\pi}{4}) = -\sqrt{2}$:$3x(\sqrt{2}) - 2y(-\sqrt{2}) = 5$$3\sqrt{2}x + 2\sqrt{2}y = 5$Thus, the equation of the normal is $3\sqrt{2}x + 2\sqrt{2}y - 5 = 0$.

The equation of the normal to the curve $4x^2 + 9y^2 = 36$ at the point where the parametric angle is $\theta = \frac{7\pi}{4}$ is

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