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Tardigrade
Question
Mathematics
The area of the quadrilateral with its vertices at the foci of the conics 9 x2-16 y2-18 x+32 y-23=0 and 25 x2+9 y2-50 x-18 y+33=0, is
Q. The area of the quadrilateral with its vertices at the foci of the conics
9
x
2
−
16
y
2
−
18
x
+
32
y
−
23
=
0
and
25
x
2
+
9
y
2
−
50
x
−
18
y
+
33
=
0
, is
1141
172
Application of Integrals
Report Error
Answer:
0.889
Solution:
1
st
is a hyperbola
9
(
x
−
1
)
2
−
16
(
y
−
1
)
2
=
16
with
e
=
5/4
and
2
nd
is an ellipse
25
(
x
−
1
)
2
+
9
(
y
−
1
)
2
=
1
with
e
=
4/5
with
x
−
1
=
X
and
y
−
1
=
Y
area
=
2
1
d
1
d
2
=
2
1
⋅
3
10
⋅
15
8
=
9
8
e
E
⋅
e
H
=
1