If the equations x = 1 + 2 cos theta and y = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given equations are $x = 1 + 2 \cos 	heta$ and $y = 2 + \sin 	heta$. This represents an ellipse centered at $(h, k) = (1, 2)$ with $a = 2$ a

Vedclass · Accepted Answer

$\left(\frac{8+\sqrt{2}}{4}, 2\right)$

Vedclass · Answer

(C) The given equations are $x = 1 + 2 \cos 	heta$ and $y = 2 + \sin 	heta$. This represents an ellipse centered at $(h, k) = (1, 2)$ with $a = 2$ and $b = 1$. The equation of the ellipse is $\frac{(x-1)^2}{4} + \frac{(y-2)^2}{1} = 1$. The parametric point $P$ at $	heta = \pi/4$ is $(1 + 2 \cos(\pi/4), 2 + \sin(\pi/4)) = (1 + \sqrt{2}, 2 + 1/\sqrt{2})$. The equation of the normal at point $(a \cos 	heta, b \sin 	heta)$ for the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $ax \sec 	heta - by \csc 	heta = a^2 - b^2$. Shifting the origin to $(1, 2)$,the normal equation is $2(x-1) \sec(\pi/4) - 1(y-2) \csc(\pi/4) = 2^2 - 1^2$. $2(x-1) \sqrt{2} - (y-2) \sqrt{2} = 3$. $2\sqrt{2}(x-1) - \sqrt{2}(y-2) = 3$. The major axis of this ellipse is the line $y = 2$. Substituting $y = 2$ into the normal equation: $2\sqrt{2}(x-1) - \sqrt{2}(2-2) = 3$. $2\sqrt{2}(x-1) = 3$. $x-1 = \frac{3}{2\sqrt{2}} = \frac{3\sqrt{2}}{4}$. $x = 1 + \frac{3\sqrt{2}}{4} = \frac{4+3\sqrt{2}}{4}$. Note: Re-evaluating the normal equation $ax \sec 	heta - by \csc 	heta = a^2 - b^2$ with $a=2, b=1, 	heta=\pi/4$: $2(x-1)\sqrt{2} - 1(y-2)\sqrt{2} = 3$. At $y=2$,$2\sqrt{2}(x-1) = 3 \implies x = 1 + \frac{3}{2\sqrt{2}} = 1 + \frac{3\sqrt{2}}{4}$. Given the options provided,there appears to be a discrepancy in the standard form. Based on the calculation,the intersection point is $(\frac{4+3\sqrt{2}}{4}, 2)$.

If the equations $x = 1 + 2 \cos \theta$ and $y = 2 + \sin \theta$ for $0 \leq \theta < 2 \pi$ represent an ellipse,then the point of intersection of the normal drawn at $P(\theta = \pi/4)$ to this ellipse and its major axis is:

Similar Questions

Let $S(1,0)$ and $S^{\prime}(0,1)$ be the foci of an ellipse such that $SP+S^{\prime} P=2$ for any point $P$ on the ellipse. If $A(x_1, y_1)$ and $A^{\prime}(x_2, y_2)$ are the end points of the major axis of this ellipse,then $x_1+x_2=$

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $36 x^{2}+4 y^{2}=144$.

If tangents are drawn from point $P(3 \sin \theta + 4 \cos \theta, 3 \cos \theta - 4 \sin \theta)$ where $\theta = \frac{\pi}{8}$ to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$,then the angle between the tangents is:

The line $y = mx + c$ is a normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,if $c = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module