AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

The given equation of the ellipse is,
‌

x2

9

+‌

y2

5

=1
The eccentricity of the ellipse is,
e=√1−‌

5

9

=‌

2

3

The coordinates of the end points of latusrectum are
L(2,‌

5

3

),M(−2,‌

5

3

),M′(−2,‌

−5

3

) and L′(2,‌

−5

3

)
The equations of tangents at these points are,
2x+3y−9=0
−2x+3y−9=0
2x+3y+9=0
−2x+3y+9=0
By solving above equations,
A(0,3)
B(−‌

9

2

,0)
C(0,−3)
D(‌

9

2

,0)
And
AC=√(3+3)2=6
BD=√(‌

9

2

+‌

9

2

)2=9
The area of the quadrilateral is,
A‌‌=‌

1

2

×9×6
‌‌=27‌ sq units ‌