Prove the following?

If #x=asin theta# and #y=bsin theta#,
Prove that #a^2/x^2-b^2/y^2=1#

1 Answer
maganbhai P.
Apr 4, 2018

Please check your equn.
#color(red)(y=bsintheta#
I think
#color(red)(y=btantheta# is O.K.

Explanation:

We have ,

#color(red)(x=asintheta and y=bsintheta#

#=>a/x=1/sintheta and b/y=1/sintheta#

Squaring both sides of both equn .

#a^2/x^2=csc^2theta and b^2/y^2=csc^2theta#

Comparing both equn. for #csc^2x#

#a^2/x^2=b^2/y^2#

#a^2/x^2-b^2/y^2=0#

So ,

#a^2/x^2-b^2/y^2!=1#

If we take,

#color(red)(x=asintheta and y=btantheta#

#=>a/x=1/sintheta=csctheta and b/y=1/tantheta=cottheta#

Squaring both sides of both equn.

#a^2/x^2=csc^2theta ...to(I)#

#b^2/y^2=cot^2theta...to(II)#

We know that,

#csc^2theta-cot^2theta=1#

Using #(I)and(II)# ,we get

#a^2/x^2-b^2/y^2=1#

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