The area of the quadrilateral formed by drawi | vedclass

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

Question

(A) For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the ends of the latus recta are $(\pm ae, \pm \frac{b^2}{a})$.Given $a^2 = 4$ and $b^2 = 1

Vedclass · Accepted Answer

$\frac{16}{\sqrt{3}}$

Vedclass · Answer

(A) For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the ends of the latus recta are $(\pm ae, \pm \frac{b^2}{a})$.Given $a^2 = 4$ and $b^2 = 1$,we have $a = 2$ and $b = 1$.The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.The coordinates of the ends of the latus recta are $(\pm \sqrt{3}, \pm \frac{1}{2})$.The equation of the tangent at $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.For $(\sqrt{3}, \frac{1}{2})$,the tangent is $\frac{x\sqrt{3}}{4} + \frac{y(1/2)}{1} = 1 \Rightarrow \frac{\sqrt{3}}{4}x + \frac{1}{2}y = 1$.The four tangents form a rhombus with vertices at $(\pm \frac{a^2}{ae}, 0) = (\pm \frac{4}{\sqrt{3}}, 0)$ and $(0, \pm \frac{b^2}{b^2/a}) = (0, \pm a) = (0, \pm 2)$.The area of the rhombus is $2 	imes 	ext{base} 	imes 	ext{height} = 2 	imes |\frac{4}{\sqrt{3}}| 	imes |2| = \frac{16}{\sqrt{3}}$.

The area of the quadrilateral formed by drawing tangents at the ends of the latus recta of the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$ is

Similar Questions

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:

The equation of the ellipse whose latus rectum is $8$ and whose eccentricity is $\frac{1}{\sqrt{2}}$,referred to the principal axes of coordinates,is

If the line $2x + 5y = 12$ intersects the ellipse $4x^2 + 5y^2 = 20$ in two distinct points $A$ and $B$,then the mid-point of $AB$ is

The length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$,whose mid-point is $(1, \frac{2}{5})$,is equal to:

The foci of the ellipse $2x^2 + 3y^2 - 4x - 12y + 13 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module