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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

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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.

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$

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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.