Question

(ax-by) varies √xy is, prove that x^2+ y^2 varies xy

Answer 1

Given

ax+by∝√xy

⇒ax+by=k√xy , where k = proportionality constant

⇒√xyax+by=1k

⇒xy(ax+by)2=1k2

⇒4abxy(ax+by)2=4abk2

⇒1−4abxy(ax+by)2=1−4abk2

⇒(ax+by)2−4abxy(ax+by)2=k2−4abk2

⇒(ax−by)2(ax+by)2=k2−4abk2=m2(say), where m= constant

⇒ax−byax+by=m

⇒ax+byax−by=1m

Now by componendo and dividendo we get

⇒2ax2by=1+m1−m

⇒x=1+m1−m×ba×y

⇒x=n×y,

where 1+m1−m×ba=n→another constant

Now

ax2+by2xy

=an2y2+by2ny2

=an2+bn→ A CONSTANT

Hence

(ax2+by2)∝xy

Answer 2

Step-by-step explanation:

see the above picture,,,,,,,,,,,,,,