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

Application Of Derivatives

Question
CBSEENMA12035146
Wired Faculty App

Find the equation of the tangent to the curve straight y equals space fraction numerator straight x minus 7 over denominator left parenthesis straight x minus 2 right parenthesis thin space left parenthesis straight x minus 3 right parenthesis end fraction at the point where  it cuts the x-axis.

Solution

The equation of curve is
                          straight y space equals space fraction numerator straight x minus 7 over denominator left parenthesis straight x minus 2 right parenthesis thin space left parenthesis straight x minus 3 right parenthesis end fraction                         ...(1)
It meets x-axis where y = 0
Putting y = 0 in (1), we get
                              fraction numerator straight x minus 7 over denominator left parenthesis straight x minus 2 right parenthesis thin space left parenthesis straight x minus 3 right parenthesis end fraction space equals space 0 space space space space space or space space space straight x minus 7 space equals space 0 space space space space or space space space straight x space equals space 7
therefore space space space space space curve space left parenthesis 1 right parenthesis space meets space straight x minus axis space in space left parenthesis 7 comma space 0 right parenthesis From space left parenthesis 1 right parenthesis comma space space space space space straight y space equals space fraction numerator straight x minus 7 over denominator left parenthesis straight x minus 2 right parenthesis thin space left parenthesis straight x minus 3 right parenthesis end fraction space equals space fraction numerator straight x minus 7 over denominator straight x squared minus 5 straight x plus 6 end fraction space space space space space space space space space space space space space space dy over dx space equals space fraction numerator left parenthesis straight x squared minus 5 straight x plus 6 right parenthesis thin space 1 minus left parenthesis straight x minus 7 right parenthesis. space left parenthesis 2 straight x minus 5 right parenthesis over denominator left parenthesis straight x squared minus 5 straight x plus 6 right parenthesis squared end fraction At space left parenthesis 7 comma space 0 right parenthesis space space dy over dx equals space fraction numerator left parenthesis 49 minus 35 plus 6 right parenthesis space minus space left parenthesis 0 right parenthesis space left parenthesis 14 minus 5 right parenthesis over denominator left parenthesis 49 minus 35 plus 6 right parenthesis squared end fraction space equals space 1 over 20 therefore space space space space the space equation space of space tangent space at space left parenthesis 7 comma space 0 right parenthesis space is space space space space space space space space space space space space space space space space space space space space space straight y minus 0 space equals 1 over 20 left parenthesis straight x minus 7 right parenthesis space space space space or space space 20 straight y space equals space straight x minus 7 space space space space or space space straight x minus 20 straight y minus 7 space equals space 0

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