The product of the perpendicular distances dr | vedclass

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

Question

(B) For the ellipse $\frac{x^2}{36}+\frac{y^2}{27}=1$,we have $a^2=36$ and $b^2=27$. The foci are $(\pm ae, 0)$.Since $e^2 = 1 - \frac{b^2}{a^2} = 1 -

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(B) For the ellipse $\frac{x^2}{36}+\frac{y^2}{27}=1$,we have $a^2=36$ and $b^2=27$. The foci are $(\pm ae, 0)$.Since $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{27}{36} = 1 - \frac{3}{4} = \frac{1}{4}$,we have $e = \frac{1}{2}$.The foci are $(\pm 6 	imes \frac{1}{2}, 0) = (\pm 3, 0)$.$A$ well-known property of an ellipse is that the product of the perpendicular distances from the foci to any tangent is equal to the square of the semi-minor axis,$b^2$.Here,$b^2 = 27$.Thus,the product of the perpendicular distances from the foci $(3,0)$ and $(-3,0)$ to the tangent at any point is $27$.

The product of the perpendicular distances drawn from the points $(3,0)$ and $(-3,0)$ to the tangent of the ellipse $\frac{x^2}{36}+\frac{y^2}{27}=1$ at the point $\left(3, \frac{9}{2}\right)$ is:

Similar Questions

The eccentric angle of a point on the ellipse $x^2+3y^2=6$ lying at a distance of $2$ units from its centre is

If tangents are drawn from any point on the circle $x^2+y^2=25$ to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$,then the angle between the tangents is

For what value of $c$ does the line $y = 4x + c$ touch the curve $\frac{x^2}{4} + y^2 = 1$? Find the number of possible values of $c$.

Find the equation for the ellipse that satisfies the given conditions: Major axis on the $x-$ axis and passes through the points $(4, 3)$ and $(6, 2)$.

Find the length of the latus rectum of the ellipse $4x^2 + 9y^2 - 36y + 4 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module