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Byju's Answer
Standard XII
Mathematics
Tangent To a Parabola
Tangents are ...
Question
Tangents are drawn to the ellipse
x
2
9
+
y
2
5
=
1
at the ends of a latus rectum. The area of the quadrilateral to formed is A. Then (A/9) is
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Solution
Ellipse:
x
2
9
+
y
2
5
=
1
(
e
=
2
3
)
end of L.R
L
:
(
2
,
5
3
)
&
L
′
(
2
,
−
5
3
)
T
be the tangent at
L
⇒
2
x
9
+
y
3
=
1
−
−
−
(
i
)
As in the intercept form so, intercept of
i
on
y
−
axis is
A
(
0
,
3
)
.
T
′
be the tangent at
L
′
=
2
x
9
+
−
y
3
=
1
−
−
−
(
i
i
)
intercept of
i
i
on
y
−
axis is
B
(
0
,
−
3
)
So the area of quadrilateral,
A
B
L
′
L
=
1
2
∣
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∣
0
3
2
5
3
2
−
5
3
0
−
3
0
3
∣
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∣
A
=
∣
∣
∣
1
2
(
0
−
6
−
10
3
−
10
3
−
6
−
0
)
∣
∣
∣
A
=
1
2
(
45
3
)
A
=
23
so,
A
9
=
23
9
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