Prove that curve(x/a)n+(y/b)n=2 touches the straight line x/a+y/b=2 at (a,b) for all values of n?
Answers
Answered by
0
Answer
Given (
a
x
)
n
+(
b
y
)
n
=2
Differentiating both sides w.r.t x, we get
a
n
(
a
x
)
n−1
+
b
b
(
b
y
)
n−1
×
dx
dy
=0
⇒
dx
dy
=−
a
n
(
a
x
)
n−1
×
a
b
(
y
b
)
n−1
∴
dx
dy
at (a,b)=
a
b
∴ Tangent is y−b=−
a
b
(x−a)⇒bx+ay=2ab⇒
a
x
+
b
y
=2
for all values of n (∵
dx
dy
is independent of n)
Answered by
0
Answer:
(a/b)c+ (c/d)n= 3 touches the straight line x/a+y/b=2 atn
Similar questions