+91-8076753736[email protected]
logo
logo
-->

Application Of Derivatives

Question
CBSEENMA12035079
Wired Faculty App

Find the equation of tangent and normal to the hyperbola straight x squared over straight a squared minus straight y squared over straight b squared space equals space 1 space space at space space left parenthesis straight x subscript 0 comma space space straight y subscript 0 right parenthesis.

Solution

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

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation