Application of Derivatives

Find the equation of the tangent and normal to the given curves at the points given: Also find the points of intersection where both tangent and normal cut the x-axis. - Wired Faculty

Question

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Find the equation of the tangent and normal to the given curves at the points given:
$square root of straight x plus square root of straight y space equals space straight a space space space at space space open parentheses straight a squared over 4 comma space space straight a squared over 4 close parentheses$
Also find the points of intersection where both tangent and normal cut the x-axis.

Solution

The equation of curve is $square root of straight x plus square root of straight y space equals space square root of straight a$
Differentiating both sides w.r.t. x, we get,.
$fraction numerator 1 over denominator 2 square root of straight x end fraction plus fraction numerator 1 over denominator 2 square root of straight y end fraction dy over dx space equals space 0 space space space space space space space space space space space space space space rightwards double arrow space space space space space dy over dx space equals space minus fraction numerator square root of straight y over denominator square root of straight x end fraction$
$At space open parentheses straight a squared over 4 comma straight a squared over 4 space close parentheses space space dy over dx space equals space fraction numerator negative square root of begin display style straight a squared over 4 end style end root over denominator square root of begin display style straight a squared over 4 end style end root end fraction space equals space minus 1 therefore space space space space space slope space of space tangent space space equals space minus 1 therefore space space space space space space the space equation space of space tangent space at space open parentheses straight a squared over 4 comma space straight a squared over 4 close parentheses space is space space space space space space space space space space space space space space space space space space straight y minus straight a squared over 4 space equals space minus open parentheses straight x minus straight a squared over 4 close parentheses space space space or space space space space straight y minus straight a squared over 4 space equals space minus straight x plus straight a squared over 4 space space space space or space space straight x plus straight y minus straight a squared over 2 space equals space 0 or space space space 2 straight x plus 2 straight y minus straight a squared space equals space 0 slope space of space normal space space equals space minus fraction numerator 1 over denominator negative 1 end fraction space equals 1 therefore space space space space space equation space of space normal space at space open parentheses straight a squared over 4 comma space space space straight a squared over 4 close parentheses space space is space space space straight y minus straight a squared over 4 space equals space 1 space open parentheses straight x minus straight a squared over 4 close parentheses space space space space or space space space space straight y minus straight a squared over 4 space equals space straight x minus straight a squared over 4 space space space or space space space straight x minus straight y space equals space 0$

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Application Of Derivatives

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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