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Application Of Derivatives

Question
CBSEENMA12035236

If the curve αx+ βy2 = 1 and α' x+ β'y2 = 1 intersect orthogonally, prove that (α – α') β β') = (β – β') α α'. 

Solution

The equations of two curves are
                            αx squared plus βy squared space equals space 1                                       ...(1)
and                 straight alpha apostrophe straight x squared plus straight beta apostrophe straight y squared space equals space 1                                          ...(2)
Let curves (1) and (2) intersect at left parenthesis straight x subscript 1 comma space space straight y subscript 1 right parenthesis.
therefore space space space space space space space space space space αx subscript 1 squared plus βy subscript 1 squared space equals space 1                                                   ...(3)
and          straight alpha apostrophe straight x subscript 1 squared plus straight beta to the power of apostrophe straight y subscript 1 squared space equals space 1                                               ...(4)
Subtracting (4) from (3), we get,
                          left parenthesis straight alpha minus straight alpha apostrophe right parenthesis space straight x subscript 1 squared space space plus space left parenthesis straight beta minus straight beta apostrophe right parenthesis straight y subscript 1 squared space equals space 0
therefore space space space space left parenthesis straight alpha minus straight a to the power of apostrophe right parenthesis space straight x subscript 1 squared space equals space minus left parenthesis straight beta minus straight beta apostrophe right parenthesis space straight y subscript 1 squared
therefore space space space space space space space space space space space space straight x subscript 1 squared over straight y subscript 1 squared space equals space fraction numerator straight beta apostrophe space minus straight beta over denominator straight alpha minus straight alpha apostrophe end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 5 right parenthesis
Differentiating space left parenthesis 1 right parenthesis space straight w. straight r. straight t. straight x comma space we space get comma
space space space space space space space 2 ax plus 2 βy dy over dx space equals space 0 space space space space space space or space space dy over dx space space equals negative αx over βy
At left parenthesis straight x subscript 1 comma space space straight y subscript 1 right parenthesis comma space space dy over dx space equals space minus αx subscript 1 over βy subscript 1
Similarly for second curve,
                            dy over dx space equals space minus fraction numerator straight alpha apostrophe straight x subscript 1 over denominator straight beta apostrophe straight y subscript 1 end fraction
Since the two curves (1) and (2) intersect orthogonally,
therefore space space space space space space space minus αx subscript 1 over βy subscript 1 cross times negative fraction numerator straight alpha apostrophe straight x subscript 1 over denominator straight beta apostrophe straight y subscript 1 end fraction space equals space minus 1 space space space space space space space space space space space space space space space space space space space space space space space open square brackets Product space of space slopes space space equals space minus 1 close square brackets
therefore space space space space space space space space space space fraction numerator αα apostrophe over denominator ββ apostrophe end fraction straight x subscript 1 squared over straight y subscript 1 squared equals negative 1
therefore space space space space space space space space space space space fraction numerator αα to the power of apostrophe over denominator ββ apostrophe end fraction cross times fraction numerator straight beta apostrophe minus straight beta over denominator straight alpha minus straight alpha apostrophe end fraction space equals space minus 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 5 right parenthesis close square brackets space
therefore space space space space αα apostrophe left parenthesis straight beta to the power of apostrophe minus straight beta right parenthesis space equals space minus left parenthesis straight alpha minus straight alpha to the power of apostrophe right parenthesis space ββ apostrophe
therefore space space space space αα apostrophe left parenthesis straight beta minus straight beta apostrophe right parenthesis space equals space left parenthesis straight alpha minus straight alpha apostrophe right parenthesis thin space ββ apostrophe
or space space space space space space space left parenthesis straight alpha minus straight alpha apostrophe right parenthesis thin space ββ apostrophe space space equals space left parenthesis straight beta minus straight beta apostrophe right parenthesis space straight alpha space straight alpha apostrophe space space space which space is space required space condition. space space