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 locus of the point of intersection of mutually perpendicular tangents to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

If $3x + 4y = 12\sqrt{2}$ is a tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{9} = 1$ for some $a \in R$,then the distance between the foci of the ellipse is:

If the latus rectum of an ellipse subtends a right angle at the centre of that ellipse,then the eccentricity of that ellipse is

Let $e$ be the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. If $a=5, b=4$ and the equation of the normal drawn at one end of the latus rectum that lies in the first quadrant is $lx+my=27$,then $l+m=$

Let the product of the focal distances of the point $\left(\sqrt{3}, \frac{1}{2}\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ be $\frac{7}{4}$. Then the absolute difference of the eccentricities of two such ellipses is

Vedclass Test Series

Exam Paper Generator

Online Exam Module