The coordinates of any point,in the parametri | vedclass

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

Question

(B) The foci are $(-2,0)$ and $(8,0)$.The distance between the foci is $2ae = 8 - (-2) = 10$.$\Rightarrow ae = 5$.Given $e = \frac{1}{\sqrt{2}}$,we ha

Vedclass · Accepted Answer

$(3+5 \sqrt{2} \cos 	heta, 5 \sin 	heta)$

Vedclass · Answer

(B) The foci are $(-2,0)$ and $(8,0)$.The distance between the foci is $2ae = 8 - (-2) = 10$.$\Rightarrow ae = 5$.Given $e = \frac{1}{\sqrt{2}}$,we have $a \left(\frac{1}{\sqrt{2}}ight) = 5 \Rightarrow a = 5\sqrt{2}$.Now,$b^2 = a^2(1 - e^2) = (5\sqrt{2})^2 \left(1 - \frac{1}{2}ight) = 50 \left(\frac{1}{2}ight) = 25$.Thus,$b = 5$.The centre of the ellipse is the mid-point of the line joining the two foci: $\left(\frac{-2+8}{2}, \frac{0+0}{2}ight) = (3,0)$.The equation of the ellipse is $\frac{(x-3)^2}{a^2} + \frac{(y-0)^2}{b^2} = 1$,which is $\frac{(x-3)^2}{50} + \frac{y^2}{25} = 1$.The parametric coordinates are $(x, y) = (h + a \cos 	heta, k + b \sin 	heta)$,where $(h, k)$ is the centre.Substituting the values,we get $(3 + 5\sqrt{2} \cos 	heta, 5 \sin 	heta)$.

The coordinates of any point,in the parametric form,on the ellipse whose foci are $(-2,0)$ and $(8,0)$ and eccentricity is $\frac{1}{\sqrt{2}}$,is

Similar Questions

In an ellipse with its centre at the origin,if the difference between the lengths of the major axis and the minor axis is $10$ and one of the foci is at $(0, 5\sqrt{3})$,then the length of its latus rectum is:

If $4x + 2y + n = 0$ is a normal to the ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$,then $n = $

If tangents are drawn to the ellipse $x^2+2y^2=2$,then the locus of the mid-points of the intercepts made by those tangents between the coordinate axes is

If tangents are drawn from the point $(2 + 13 \cos \theta, 3 + 13 \sin \theta)$ to the ellipse $\frac{(x-2)^2}{25} + \frac{(y-3)^2}{144} = 1$,then the angle between them is:

Find the condition for the line $ax + by + c = 0$ to be a normal to an ellipse $\frac{x^2}{4} + \frac{y^2}{36} = 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module