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Byju's Answer
Standard XII
Mathematics
Tangent of a Curve y =f(x)
P1 : y2 = 4ax...
Question
P
1
:
y
2
=
4
a
x
,
P
2
:
y
2
=
−
4
a
x
L
:
y
=
x
Equation of the tangent at a point on the parabola
P
1
where the line
L
meets the parabola is
A
x
−
2
y
+
4
a
=
0
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B
x
+
2
y
−
4
a
=
0
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C
x
+
2
y
−
8
a
=
0
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D
x
−
2
y
+
8
a
=
0
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Solution
The correct option is
A
x
−
2
y
+
4
a
=
0
Clearly
P
1
and
L
meets at
(
4
a
,
4
a
)
Thus equation of tangent at this point is given by,
y
(
4
a
)
=
2
a
(
x
+
4
a
)
⇒
x
−
2
y
+
4
a
=
0
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0
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Standard XII Mathematics
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