If $$x/aCOS \alpha+y/bSIN \alpha =1 and x/aCOS \beta+y/bSIN \beta = 1$$
and $$COS \alpha COS \beta /a ^{2}+SIN \alpha SIN \beta/b ^{2} = 0$$
Then prove that
a)$$x ^{2}+y ^{2}=a ^{2}+b ^{2}$$ and b) $$\tan \alpha \tan \beta = b ^{2}(x ^{2}-a ^{2})/a ^{2}(y ^{2}-b ^{2})$$

Question

If $$x/aCOS \alpha+y/bSIN \alpha =1 and x/aCOS \beta+y/bSIN \beta = 1$$
and $$COS \alpha COS \beta /a ^{2}+SIN \alpha SIN \beta/b ^{2} = 0$$ 
Then prove that 
a)$$x ^{2}+y ^{2}=a ^{2}+b ^{2}$$ and b) $$\tan \alpha \tan \beta = b ^{2}(x ^{2}-a ^{2})/a ^{2}(y ^{2}-b ^{2})$$

anonymous · Answer

Can somebody help me solve this???