A point on the curve x=3 cos theta, y=2 sin t | vedclass

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(A) The given curve is $x=3 \cos 	heta$ and $y=2 \sin 	heta$. This represents an ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$. $A$ tangent is perpend

Vedclass · Accepted Answer

$(3,0)$

Vedclass · Answer

(A) The given curve is $x=3 \cos 	heta$ and $y=2 \sin 	heta$. This represents an ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$. $A$ tangent is perpendicular to the $X$-axis when the slope of the tangent is undefined,which occurs when $\frac{dx}{d	heta} = 0$ and $\frac{dy}{d	heta} 
eq 0$. Calculating the derivatives: $\frac{dx}{d	heta} = -3 \sin 	heta$ and $\frac{dy}{d	heta} = 2 \cos 	heta$. Setting $\frac{dx}{d	heta} = 0$ gives $-3 \sin 	heta = 0$,so $\sin 	heta = 0$,which means $	heta = 0$ or $	heta = \pi$. For $	heta = 0$,$x = 3 \cos(0) = 3$ and $y = 2 \sin(0) = 0$. For $	heta = \pi$,$x = 3 \cos(\pi) = -3$ and $y = 2 \sin(\pi) = 0$. Thus,the points are $(3, 0)$ and $(-3, 0)$. Comparing with the given options,$(3, 0)$ is the correct point.

$A$ point on the curve $x=3 \cos \theta, y=2 \sin \theta$ at which the tangent is perpendicular to the $X$-axis is

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