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Application Of Derivatives

Question
CBSEENMA12035232

Prove that the curves x = y2 and xy = k cut at right angles if 8k2 = 1.

Solution

The equation of two curves are
                           x = y2                                                                            ...(1)
             and       xy = k                                                                              ...(2)
From (1) and (2), straight y squared. space straight y space equals space straight k space space space space space space space rightwards double arrow space space straight y cubed space equals space straight k space space space rightwards double arrow space space space space straight y space equals space straight k to the power of 1 third end exponent
therefore space space space space space from space left parenthesis 1 right parenthesis comma space space straight x space equals space straight k to the power of 2 over 3 end exponent
therefore space space space space space curves space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis space intersect space at space open parentheses straight k to the power of 2 over 3 end exponent comma space space space straight k to the power of 1 third end exponent close parentheses
space space space space From space left parenthesis 1 right parenthesis comma space space space space 1 space equals space 2 straight y dy over dx space space space space space space space space space space space space space space rightwards double arrow space space space space dy over dx space equals space fraction numerator 1 over denominator 2 straight y end fraction
space space At space open parentheses straight k to the power of 2 over 3 end exponent comma space space straight k to the power of 1 third end exponent close parentheses comma space space dy over dx space equals space fraction numerator 1 over denominator 2 straight k to the power of begin display style 1 third end style end exponent end fraction
therefore space space space space space straight m subscript 1 space equals space fraction numerator 1 over denominator 2 straight k to the power of begin display style 1 third end style end exponent end fraction space where space straight m subscript 1 space is space slope space of space tangent space to space the space curve space left parenthesis 1 right parenthesis space at space the space point space of space intersection.
From (2),  straight x dy over dx plus straight y space equals space 0 space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space space space dy over dx space equals space minus straight y over straight x
At open parentheses straight k to the power of 2 over 3 end exponent comma space space space straight k to the power of 1 third end exponent close parentheses comma space space space dy over dx space equals space minus straight k to the power of begin display style 1 third end style end exponent over straight k to the power of begin display style 2 over 3 end style end exponent space equals space minus 1 over straight k to the power of begin display style 1 third end style end exponent
therefore space space space space space straight m subscript 2 space equals space minus 1 over straight k to the power of begin display style 1 third end style end exponent space where space straight m subscript 2 space is space slope space of space tangent space to space the space curve space left parenthesis 2 right parenthesis space at space the space point space of space intersection.
Curves (1) and (2) cut at right angles if straight m subscript 1 straight m subscript 2 space equals space minus 1
i.e.,  if space fraction numerator 1 over denominator 2 straight k to the power of begin display style 1 third end style end exponent end fraction. space space fraction numerator negative 1 over denominator straight k to the power of begin display style 1 third end style end exponent end fraction space equals space minus 1 space space space space straight i. straight e. space space if space space space space 2 straight k to the power of 2 over 3 end exponent space equals space 1 space space space space straight i. straight e. space space if space space 8 straight k squared space equals space 1
Hence the result. 

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