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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.

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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.

The equation of given curve is $straight y space equals space be to the power of negative straight x over straight a end exponent$ ...(1)
It cuts y-axis where x = 0
$therefore space space space space space space putting space straight x space equals space 0 space in space left parenthesis 1 right parenthesis comma space we space get comma space space space space space space space space space space space space straight y equals be to the power of 0 space equals space straight b cross times 1 space equals space straight b therefore space space space space space curve space left parenthesis 1 right parenthesis space cuts space straight y minus axis space at space left parenthesis 0 comma space straight b right parenthesis.$
Differentiating (1), w.r.t. x, we get,
$dy over dx space equals space minus straight b over straight a straight e to the power of negative straight x over straight a end exponent$

$At space left parenthesis 0 comma space straight b right parenthesis comma space space dy over dx space equals space minus straight b over straight a straight e to the power of 0 space equals space minus straight b over straight a cross times 1 space equals space minus straight b over straight a therefore space space space space straight m space equals space minus straight b over straight a$
The equation of tangent (0, b) with $straight m space equals space minus straight b over straight a space space is$
$straight y minus straight b space equals space minus straight b over straight a left parenthesis straight x minus 0 right parenthesis$ $open square brackets straight y minus straight y subscript 1 space equals space straight m left parenthesis straight x minus straight x subscript 1 right parenthesis close square brackets$

$space space space space space ay minus ab space equals space minus bx comma space space space space space or space space space bx plus ay space equals space ab space or space space space space space straight x over straight a plus straight y over straight b space space equals space 1 Hence space the space result. space$