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Tardigrade
Question
Mathematics
Area of the triangle formed by the vertex, focus and one end of latusrectum of the parabola (x+2)2=-12(y-1) is
Q. Area of the triangle formed by the vertex, focus and one end of latusrectum of the parabola
(
x
+
2
)
2
=
−
12
(
y
−
1
)
is
6
0
Conic Sections
Report Error
A
36
B
18
C
9
D
6
Solution:
Consider equal parabola
y
2
=
12
x
.
Vertex:
V
(
0
,
0
)
Focus:
S
(
3
,
0
)
End point of Latus Rectum:
A
(
3
,
6
)
∴
Required area
=
2
1
S
V
×
A
S
=
9
sq. units