The area of the quadrilateral formed by the t | vedclass

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

Question

(D) For the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$,we have $a^2 = 9$ and $b^2 = 5$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(D) For the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$,we have $a^2 = 9$ and $b^2 = 5$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.The coordinates of the end points of the latus rectum are $(\pm ae, \pm \frac{b^2}{a}) = (\pm 2, \pm \frac{5}{3})$.The tangent at the point $(2, \frac{5}{3})$ is $\frac{2x}{9} + \frac{5y/3}{5} = 1$,which simplifies to $\frac{2x}{9} + \frac{y}{3} = 1$,or $\frac{x}{9/2} + \frac{y}{3} = 1$.This tangent forms a triangle with the coordinate axes in the first quadrant with intercepts $x_0 = 9/2$ and $y_0 = 3$.The area of this triangle is $\frac{1}{2} 	imes \frac{9}{2} 	imes 3 = \frac{27}{4}$.Since there are four such symmetric triangles formed by the tangents at the four ends of the latus recta,the total area of the quadrilateral is $4 	imes \frac{27}{4} = 27$ $sq. 	ext{ units}$.

The area of the quadrilateral formed by the tangents at the end points of the latus rectum to the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$ is .............. $sq. \text{ units}$.

Similar Questions

The eccentricity of the ellipse $(x - 3)^2 + (y - 4)^2 = \frac{y^2}{9}$ is

Find the number of real tangents that can be drawn from the point $(3, 5)$ to the ellipses $3x^2 + 5y^2 = 32$ and $25x^2 + 9y^2 = 450$.

The eccentricity of the ellipse $25x^2 + 16y^2 - 150x - 175 = 0$ is

The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(a \cos \theta, b \sin \theta)$ is

The normal at a variable point $P$ on an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ of eccentricity $e$ meets the axes of the ellipse in $Q$ and $R$. Then the locus of the mid-point of $QR$ is a conic with an eccentricity $e'$ such that:

Vedclass Test Series

Exam Paper Generator

Online Exam Module