The coordinates of a point,in the parametric | vedclass

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

Question

(B) The center of the ellipse is the midpoint of the foci $(-1, 0)$ and $(7, 0)$,which is $(\frac{-1+7}{2}, \frac{0+0}{2}) = (3, 0)$.The distance betw

Vedclass · Accepted Answer

$(3 + 8 \cos 	heta, 4 \sqrt{3} \sin 	heta)$

Vedclass · Answer

(B) The center of the ellipse is the midpoint of the foci $(-1, 0)$ and $(7, 0)$,which is $(\frac{-1+7}{2}, \frac{0+0}{2}) = (3, 0)$.The distance between the foci is $2ae = 7 - (-1) = 8$,so $ae = 4$.Given $e = \frac{1}{2}$,we have $a(\frac{1}{2}) = 4$,which implies $a = 8$.Using the relation $b^2 = a^2(1 - e^2)$,we get $b^2 = 8^2(1 - (\frac{1}{2})^2) = 64(1 - \frac{1}{4}) = 64(\frac{3}{4}) = 48$.Thus,$b = \sqrt{48} = 4 \sqrt{3}$.The parametric coordinates of a point on an ellipse with center $(h, k)$ are $(h + a \cos 	heta, k + b \sin 	heta)$.Substituting the values,we get $(3 + 8 \cos 	heta, 0 + 4 \sqrt{3} \sin 	heta) = (3 + 8 \cos 	heta, 4 \sqrt{3} \sin 	heta)$.

The coordinates of a point,in the parametric form,on the ellipse whose foci are $(-1, 0)$ and $(7, 0)$ and eccentricity $e = \frac{1}{2}$,are

Similar Questions

The eccentricity of the ellipse $9x^2 + 5y^2 - 18x - 2y - 16 = 0$ is

For $k>0$,the shortest distance from a point $P(1, k)$ on the ellipse $9x^2+4y^2-18x+16y-11=0$ to one of its directrices is

The eccentricity of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is:

The minimum distance between two points $P$ and $Q$ on the ellipse $\frac{x^2}{25} + \frac{y^2}{4} = 1$,if the difference between the eccentric angles of $P$ and $Q$ is $\frac{3\pi}{2}$,is

Let $P(2,2)$ be a point on an ellipse whose foci are $F_{1}(5,2)$ and $F_{2}(2,6)$. Find the eccentricity of the ellipse.

Vedclass Test Series

Exam Paper Generator

Online Exam Module