Question

IF x square+5ysquare varies to 3xy.
Prove that x to the power 4 +y to the power 4 varies to x cube y


Please give me the solution​

Answer 1

Answer:

Let x=ky then,

x

2

+y

2

=k

2

y

2

+y

2

=y

2

(k

2

+1)

x

2

−y

2

=k

2

y

2

−y

2

=y

2

(k

2

−1)

So

x

2

−y

2

x

2

+y

2

=

(k

2

−1)

(k

2

+1)

x

2

+y

2

=l(x

2

−y

2

) where l=

(k

2

−1)

(k

2

+1)

, a constant.

Hence proved.

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Answer 2

Answer:

Assuming 2x2+3y2=5kxy ….(1), k=constant is what it’s given,

First method is since (1) is homogeneous equation, y=mx is its solution, putting which in (1) gives the equation for m in terms of k 3m2–5km+2=0 or m=5k+−25k2–24−−−−−−−√6 . Now, x+yx−y=1+(y/x)1−(y/x)=1+m1−m = another constant, so we can say that x+y varies as x-y. This is sufficient to prove that is required

I first did this and then I remembered the part 1. and the result is unexpected. I got the value of m!!Divide (1) by y2,2(x/y)2+3=5k(x/y) …(2) Divide (1) by x2,2+3(y/x)2=5k(y/x) …(3) Solving (2) and (3) for x/y and y/x respectively, we get x/y=5k+−25k2–24−−−−−−−√4 …(4) and y/x=5k+−25k2–24−−−−−−−√6 …(5). From (4) and (5), 5k+−25k2–24−−−−−−−√4=65k+−25k2–24−−−−−−−√ or 24=(5k+−25k2–24−−−−−−−√)2 which gives value of (5k+−25k2–24−−−−−−−√)=+−24−−√ giving x/y=+−24−−√/4=+−6–√/2 which gives value of m of (1) as +−2/6–√