the area of a quadrilateral formed by the normal at the end points of the latusrectum of the ellipse x2/a2 +y2/b2=1 is
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Answer:
Given equation of ellipse is
9
x
2
+
5
y
2
=1 ......(1)
∴a
2
=9,b
2
=5
⇒a=3,b=
5
Now, e=
1−
a
2
b
2
=
1−
9
5
=
3
2
Foci=(±ae,0)=(±2,0)
and
a
b
2
=
3
5
∴ Extremities of one of latus rectum are (2,
3
5
) and (2,
3
−5
)
∴ Equation of tangent at (2,
3
5
) is
9
x(2)
+
5
y(
3
5
)
=1
or 2x+3y=9 .......(2)
Eqn(2) intersects X and Y axes at (
2
9
,0) and (0,3) respectively.
∴ Area of quadrilateral 4×Area of △POQ
=4×
2
1
×
2
9
×3=27sq.units.
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