The area (in square units) of the quadrilater | vedclass

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

Question

(A) Given the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$,we have $a^2 = 9$ and $b^2 = 5$,so $a = 3$ and $b = \sqrt{5}$.The eccentricity $e = \sqrt{1

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(A) Given the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$,we have $a^2 = 9$ and $b^2 = 5$,so $a = 3$ and $b = \sqrt{5}$.The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.The foci are $(\pm ae, 0) = (\pm 3 	imes \frac{2}{3}, 0) = (\pm 2, 0)$.The end points of the latera recta are $(\pm 2, \pm \frac{b^2}{a}) = (\pm 2, \pm \frac{5}{3})$.The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.For the point $(2, \frac{5}{3})$,the tangent is $\frac{x(2)}{9} + \frac{y(5/3)}{5} = 1$ $\Rightarrow \frac{2x}{9} + \frac{y}{3} = 1$ $\Rightarrow 2x + 3y = 9$.This line intersects the x-axis at $R(\frac{9}{2}, 0)$ and the y-axis at $Q(0, 3)$.The quadrilateral is formed by four such symmetric triangles in each quadrant.The area of the triangle in the first quadrant with vertices $(0, 0)$,$(0, 3)$,and $(\frac{9}{2}, 0)$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{9}{2} 	imes 3 = \frac{27}{4}$.Since there are four such congruent triangles,the total area of the quadrilateral is $4 	imes \frac{27}{4} = 27$ square units.

The area (in square units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$ is:

Similar Questions

The length of the latus rectum of the ellipse $5x^2 + 9y^2 = 45$ is

Let the equations of two ellipses be $E_1: \frac{x^2}{3} + \frac{y^2}{2} = 1$ and $E_2: \frac{x^2}{16} + \frac{y^2}{b^2} = 1$. If the product of their eccentricities is $\frac{1}{2}$,then the length of the minor axis of ellipse $E_2$ is

The normal at a point $P$ on the ellipse $x^2+4y^2=16$ meets the $x$-axis at $Q$. If $M$ is the midpoint of the line segment $PQ$,then the locus of $M$ intersects the latus rectums of the given ellipse at the points

If the line $x \cos \alpha + y \sin \alpha = p$ is normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,then

The number of values of $c$ such that the line $y=4x+c$ touches the curve $\frac{x^{2}}{4}+y^{2}=1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module