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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Differentiation
JEE Main · Mathematics · Differentiation
If x= 1-t^ 2 1+t^ 2 and y= 2 a t 1+t^ 2 , then (d y)/(d x) is equal...
If
x
=
1
−
t
2
1
+
t
2
x=\frac{1-t^{2}}{1+t^{2}}
x
=
1
+
t
2
1
−
t
2
and
y
=
2
a
t
1
+
t
2
y=\frac{2 a t}{1+t^{2}}
y
=
1
+
t
2
2
a
t
, then
d
y
d
x
\frac{d y}{d x}
d
x
d
y
is equal to
Options
A
a
(
1
−
t
2
)
2
t
\frac{a\left(1-t^{2}\right)}{2 t}
2
t
a
(
1
−
t
2
)
B
a
(
t
2
−
1
)
2
t
\frac{a\left(t^{2}-1\right)}{2 t}
2
t
a
(
t
2
−
1
)
C
a
(
t
2
+
1
)
2
t
\frac{a\left(t^{2}+1\right)}{2 t}
2
t
a
(
t
2
+
1
)
D
\frac{a
\left(t^{2}-1\right)}{t}
Correct answer
B.
a
(
t
2
−
1
)
2
t
\frac{a\left(t^{2}-1\right)}{2 t}
2
t
a
(
t
2
−
1
)
Step-by-step solution
Given,
x
=
1
−
t
2
1
+
t
2
x=\frac{1-t^{2}}{1+t^{2}}
x
=
1
+
t
2
1
−
t
2
and
y
=
2
a
t
1
+
t
2
y=\frac{2 a t}{1+t^{2}}
y
=
1
+
t
2
2
a
t
On differentiating w.r.t.
t
t
t
, respectively, we get
d
x
d
t
=
(
1
+
t
2
)
(
0
−
2
t
)
−
(
1
−
t
2
)
(
0
+
2
t
)
(
1
+
t
2
)
2
=
−
4
t
(
1
+
t
2
)
2
\begin{aligned}\frac{d x}{d t}&=\frac{\left(1+t^{2}\right)(0-2 t)-\left(1-t^{2}\right)(0+2 t)}{\left(1+t^{2}\right)^{2}}\\ &=\frac{-4 t}{\left(1+t^{2}\right)^{2}} \end{aligned}
d
t
d
x
=
(
1
+
t
2
)
2
(
1
+
t
2
)
(
0
−
2
t
)
−
(
1
−
t
2
)
(
0
+
2
t
)
=
(
1
+
t
2
)
2
−
4
t
and
d
y
d
t
=
(
1
+
t
2
)
2
a
−
2
a
t
(
2
t
)
(
1
+
t
2
)
2
=
2
a
(
1
−
t
2
)
(
1
+
t
2
)
2
∴
d
y
d
x
=
d
y
/
d
t
d
x
/
d
t
=
a
(
1
−
t
2
)
−
2
t
⇒
d
y
d
x
=
a
(
t
2
−
1
)
2
t
\frac{d y}{d t}=\frac{\left(1+t^{2}\right) 2 a-2 a t(2 t)}{\left(1+t^{2}\right)^{2}}=\frac{2 a\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}\therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a\left(1-t^{2}\right)}{-2 t}\Rightarrow \quad \frac{d y}{d x}=\frac{a\left(t^{2}-1\right)}{2 t}
d
t
d
y
=
(
1
+
t
2
)
2
(
1
+
t
2
)
2
a
−
2
a
t
(
2
t
)
=
(
1
+
t
2
)
2
2
a
(
1
−
t
2
)
∴
d
x
d
y
=
d
x
/
d
t
d
y
/
d
t
=
−
2
t
a
(
1
−
t
2
)
⇒
d
x
d
y
=
2
t
a
(
t
2
−
1
)
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