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Application Of Derivatives

Question
CBSEENMA12035077
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Find the equation of tangent to the ellipse straight x squared over straight a squared plus straight y squared over straight b squared space equals space 1 space space space at space left parenthesis straight x subscript 1 comma space space straight y subscript 1 right parenthesis.

Solution

The equation of ellipse is straight x squared over straight a squared plus straight y squared over straight b squared space equals space 1
Differentiating both sides w.r.t.x,
                fraction numerator 2 straight x over denominator straight a squared end fraction plus fraction numerator 2 straight y over denominator straight b squared end fraction dy over dx space equals space 0 space space space space space space space space or space space space fraction numerator 2 straight y over denominator straight b squared end fraction dy over dx space equals space minus fraction numerator 2 straight x over denominator straight a squared end fraction
therefore space space space space dy over dx equals negative straight b squared over straight a squared straight x over straight y
space At space left parenthesis straight x subscript 1 comma space space straight y subscript 1 right parenthesis. space space dy over dx space equals space minus straight b squared over straight a squared straight x subscript 1 over straight y subscript 1 comma space which space is space slope space of space tangent space therefore space space space space space the space equation space of space tangent space at space left parenthesis straight x subscript 1 comma space space straight y subscript 1 right parenthesis space is space space space space space space space space space space space space space space straight y minus straight y subscript 1 space equals space minus straight b squared over straight a squared straight x subscript 1 over straight y subscript 1 left parenthesis straight x subscript 1 comma space space straight y subscript 1 right parenthesis space is or space space space space straight y subscript 1 over straight b squared left parenthesis straight y minus straight y subscript 1 right parenthesis space equals space minus straight x subscript 1 over straight a squared left parenthesis straight x minus straight x subscript 1 right parenthesis space space space space or space space space space space yy subscript 1 over straight b squared minus straight y subscript 1 squared over straight a squared space equals space minus xx subscript 1 over straight a squared plus straight x subscript 1 squared over straight a squared or space space space space space xx subscript 1 over straight a squared plus yy subscript 1 over straight b squared space equals space straight x subscript 1 squared over straight a squared plus straight y subscript 1 squared over straight b squared space space space space space space space space or space space space space space space space xx subscript 1 over straight a squared plus yy subscript 1 over straight b squared space equals space 1
                                                                        open square brackets table row cell because space space left parenthesis straight x subscript 1 comma space space straight y subscript 1 right parenthesis space lies space on space straight x squared over straight a squared plus straight y squared over straight b squared space equals space 1 end cell row cell therefore space straight x subscript 1 squared over straight a squared plus fraction numerator space straight y subscript 1 squared over denominator straight b squared end fraction equals space 1 end cell end table close square brackets

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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