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

The equation of curve is $straight y space equals space fraction numerator 5 straight x squared over denominator 1 plus straight x squared end fraction$

$therefore space space space space dy over dx space equals space fraction numerator left parenthesis 1 plus straight x squared right parenthesis. space space 10 straight x minus space 5 straight x squared.2 straight x over denominator left parenthesis 1 plus straight x squared right parenthesis squared end fraction space equals space fraction numerator 10 straight x plus 10 straight x cubed minus 10 straight x cubed over denominator left parenthesis 1 plus straight x squared right parenthesis squared end fraction space equals space fraction numerator 10 straight x over denominator left parenthesis 1 plus straight x squared right parenthesis squared end fraction$

$At space left parenthesis 2 comma space 4 right parenthesis comma space space dy over dx space equals space fraction numerator 10 space cross times space 2 over denominator left parenthesis 1 plus 4 right parenthesis squared end fraction space equals space 20 over 25 space equals space 4 over 5 comma space space which space is space slope space of space tangent. therefore space space space space space space the space equation space of space tangent space at space left parenthesis 2 comma space 4 right parenthesis space is space space space space space space space straight y minus 4 space equals space 4 over 5 left parenthesis straight x minus 2 right parenthesis comma space space space or space space space 5 straight y minus 20 space equals space 4 straight x minus 8 space space space or space space 4 straight x minus 5 straight y plus 12 space equals space 0 Slope space of space normal space space equals space minus 5 over 4 therefore space space space space space space the space equation space of space normal space at space left parenthesis 2 comma space 4 right parenthesis space is space space space space space space space space space space straight y minus 4 space equals space minus 5 over 4 left parenthesis straight x minus 2 right parenthesis space comma space space space space or space space 4 straight y minus 16 space equals negative 5 straight x plus 10 or space space space space 5 straight x plus 4 straight y minus 26 space equals space 0$

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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

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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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