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Application of Derivatives

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Short Answer Type

91. Find the equation of the tangent and normal to the given curves at the points given:
y = x⁴ – 6x³ + 13x² – 10x + 5 at ( 0 , 5 )

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92. Find the equation of the tangent and normal to the given curves at the points given:
y = x⁴ –6x³+ 13x²– 10x + 5 at (1 ,3)

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93. Find the equation of the tangent and normal to the given curves at the points given:

straight y squared space equals space fraction numerator straight x cubed over denominator 4 minus straight x end fraction space space at space space left parenthesis 2 comma space space minus 2 right parenthesis

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94. Find the equation of the tangent and normal to the given curves at the points given:

straight y space equals fraction numerator 5 straight x squared over denominator 1 plus straight x squared end fraction

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Long Answer Type

95. Find the equation of the tangent and normal to the given curves at the points given:
$square root of straight x plus square root of straight y space equals space 5$

The equation of curve is
$square root of straight x plus square root of straight y space equals space 5$
Differentiating both sides w.r.t x,
$fraction numerator 1 over denominator 2 square root of straight x end fraction plus fraction numerator 1 over denominator 2 square root of straight y end fraction dy over dx space equals space 0 comma space space space space space space or space space space space fraction numerator 1 over denominator 2 square root of straight x end fraction dy over dx space equals negative fraction numerator 1 over denominator 2 square root of straight x end fraction$
$therefore space space space space dy over dx space equals space minus fraction numerator square root of straight y over denominator square root of straight x end fraction At space left parenthesis 9 comma space 4 right parenthesis comma space space space dy over dx space equals negative fraction numerator square root of 4 over denominator square root of 9 end fraction space equals space minus 2 over 3 comma space space space which space is space slope space of space tangent. therefore space space space space space space the space equation space of space tangent space at space left parenthesis 9 comma space 4 right parenthesis space is space space space space space space space space space space space space space space space space space space space space space straight y minus 4 space equals space minus 2 over 3 left parenthesis straight x minus 9 right parenthesis comma space space space space or space space 3 straight y minus 12 space equals space minus 2 straight x plus 18 space space space or space space space space 2 straight x plus 3 straight y minus 30 space equals space 0 Slope space of space normal space space equals space 3 over 2 therefore space space space space the space equation space of space normal space at space left parenthesis 9 comma space 4 right parenthesis space is space space space space space space space space space space space space space space space space straight y minus 4 space equals space 3 over 2 left parenthesis straight x minus 9 right parenthesis comma space space space space or space space 2 straight y minus 8 space equals space 3 straight x minus 27 space space space or space space space 3 straight x minus 2 straight y minus 19 space equals space 0$

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96. Find the equation of the tangent and normal to the given curves at the points given:

16 straight x squared plus 9 straight y squared space equals space 144 space space space at space left parenthesis straight x subscript 1 comma space space straight y subscript 1 right parenthesis space space space where space straight x subscript 1 space equals space 2 space space space and space straight y subscript 1 greater than 0

Also find the points of intersection where both tangent and normal cut the x-axis.

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97. 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.

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98. Find the equation of the tangent to the curve y = (x³ – 1) (x – 2) at points where the curve cuts the x-axis.

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Short Answer Type

99. 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.

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Long Answer Type

100.

Prove that the line $straight x over straight a plus straight y over straight b space equals space 1$ is a tangent to the curve $straight y space equals be to the power of negative straight x over straight a end exponent$ at the point where the curve cuts y-axis.

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