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Parametric equation of Ellipse is considered one the most difficult concept.
51 Questions around this concept.
Consider an ellipse, whose centre is at the origin and its major axis is along the $x$-axis. If its eccentricity is $\frac{3}{5}$ and the distance between its foci is 6 , then the area (in sq. units) of the quadrilateral inscribed in the ellipse, with the vertices as the end points of major and minor axes of ellipse, is
The equation of the ellipse whose axes are the axes of coordinates and which passes through the point (–3, 1) and has eccentricity is
The equation of the circle passing through the foci of the ellipse , and having centre at
(0, 3) is :
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If O is the origin and OB is the semi-minor axis of an ellipse, F1 and F2 are its foci and the angle between F1B and F2B is a right angle, then the square of the eccentricity of the ellipse is
The locus of mid-points of the line segments joining $(-3,-5)$ and the points on the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$
The foci of the ellipse are:
Parametric equation of Ellipse:
The equations $\mathrm{x}=\mathrm{a} \cos \theta, \mathrm{y}=\mathrm{b} \sin \theta$ are called the parametric equation of the ellipse.
The circle described on the major axis as the diameter is called the auxiliary circle.
Equation of Ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Then, equation of auxiliary circle is $x^2+y^2=a^2$ (As $A A^{\prime}$ is Diameter)
Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be a point on the ellipse
Draw PN perpendicular to the major axis and PN to meet the auxiliary circle at Q.
Let $\angle \mathrm{ACQ}$ be $\theta$ (This angle is also known as Eccentric Angle). Hence, the parametric equation of circle at point Q (a $\cos \Theta, \mathrm{a} \sin \Theta$ ).
Thus, P has x -coordinate as a $\cos \Theta$
As P lies on the ellipse
$
\frac{\mathrm{a}^2 \cos ^2 \theta}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \Rightarrow \mathrm{y}= \pm \mathrm{b} \sin \theta
$
Hence, Point P is $(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)$
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