The area of the quadrilateral formed by the tangents at the endpoint of the latuc rectum to the ellipse `(x^2)/9+(y0^2)/5=1` is 27/4 sq. unit (b) 9 sq. units 27/2 sq. unit (d) 27 sq. unit

Options

A. `27//4` sq units
B. 9 sq units
C. `27//2` sq units
D. 27 sq units

Answer

D

Explanation

Given, `(x^(2))/(9)+(y^(2))/(5)=1`
To find tangents at the end points of latusrectum, we find ae.
i.e. ae`=sqrt(a^(2)-b^(2))=sqrt(4)=2`
and `sqrt(b^(2)(1-e^(2)))=sqrt(5(1-(4)/(9)))=(5)/(3)`
By symmetry,the quadrilateral is a rhombus.

So, area is four times the area of the right angled triangle
formed by the tangent and axes in the Ist quadrant.
` therefore " Equation of tangent at " (2,(5)/(3)) " is"`
`(2)/(9)x+(5)/(3)*(y)/(5)=1 rArr (x)/(9//2)+(y)/(3)=1 `
` therefore " Area of quadrilateral " ABCD `
`=4 " "["area of " triangle AOB]`
`= 4((1)/(2)*(9)/(2)*3)=27 ` sq units