find the envelope of x^2/ a^2+ y^2/b^2=1 , where the parameter a are connected a^n+b^n=c^n, where c is a constant
Answers
Answer:
Step-by-step explanation:
Let the equation of straight line be
a
x
+
b
y
=1 ....(1)
Given a+b=4 ....(2)
Differentiating eqn (1) w.r.t. a , we get
−
a
2
x
−
b
2
y
da
db
=0
⇒
a
2
x
+
b
2
y
da
db
=0 ....(3)
Differentiating (2) w.r.. t a, we get
1+
da
db
=0
⇒
da
db
=−1 .....(4)
From (3) and (4), we get
a
2
x
=
b
2
y
....(5)
⇒
b
y
a
x
=
b
a
Adding 1 to both sides, we get
⇒
b
y
1
=
b
4
⇒b
2
=4y
⇒b=2
y
Also, by eqn (5), we get
a=2
x
Now, using (2), the envelope is
x
+
y
=2
Answer:find the envelope of x^2/ a^2+ y^2/b^2=1 , where the parameter a are connected a^n+b^n=c^n, where c is a constant is x/a+y/b+z/c=1
Step-by-step explanation:
Let the equation of straight line be
ax+ by=1 ....(1)
Given a+b=4 ....(2)
Differentiating eqn (1) w.r.t. a , we get
−
a 2x− b 2adb=0
⇒ a2x+ b 2y
Now solving this differntial and getting the account is the same as solving:-
to get the final solutions and the envelope of x^2/ a^2+ y^2/b^2=1 at the given constraint of the ellipse that:-
So we know how to get the equation of the envelope that:-
3.1 Rolls Parallel to a Lateral Wall
In the direction of the pattern's wavevector, the envelope equation defines a well-posed second-order differential problem if we have one condition at each end of the interval corresponding to the width of the system. At lowest order we usually have
(11)
or, for a semi-infinite system,
Writing A(x) = |A(x)|exp(i ϕ(x)), we can get the steady solution explicitly; the equation for ϕ reads
which yields:-
So using this theoram we get the equation of the envelope as:-
Answer:-find the envelope of x^2/ a^2+ y^2/b^2=1 , where the parameter a are connected a^n+b^n=c^n, where c is a constant is x/a+y/b+z/c=1.
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