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Application of Derivatives

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Short Answer Type

61.

Find the slope of the normal at the point (am³ , am² ) to the curve a x² = y³.

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62.

Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).

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63.

Find the point at which the tangent to the curve $straight y equals space square root of 4 straight x minus 3 end root space minus 1$ has its slope $2 over 3.$

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64.

Find the equation of the normal to the curve y² = 4x at the point (1, 2).

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65.

Find the equation of tangent to the ellipse $straight x squared over straight a squared plus straight y squared over straight b squared space equals space 1 space space space at space left parenthesis straight x subscript 1 comma space space straight y subscript 1 right parenthesis.$

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66.

Find the equation of tangent and normal to the hyperbola $straight x squared over straight a squared minus straight y squared over straight b squared space equals space 1 space space at space space left parenthesis straight x subscript 0 comma space space straight y subscript 0 right parenthesis.$

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67. Find the equation of the normal at the point (am², am³) for the curve ay²= x³.

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68.
Find the equation of the tangent and the normal to the curve $straight x to the power of 2 over 3 end exponent plus straight y to the power of 2 over 3 end exponent space equals space 2 space space at space left parenthesis 1 comma space 1 right parenthesis$

The equation of curve is
$straight x to the power of 2 over 3 end exponent plus straight y to the power of 2 over 3 end exponent space equals space 2$
Differentiating both sides w.r.t.x, we get.
$2 over 3 straight x to the power of negative 1 third end exponent plus 2 over 3 straight y to the power of negative 1 third end exponent dy over dx space equals space 0 space space space space space space space or space space space space straight y to the power of negative 1 third end exponent dy over dx space equals space minus straight x to the power of negative 1 third end exponent$
$therefore space space space space space space space dy over dx space equals space minus straight x to the power of negative begin display style 1 third end style end exponent over straight y to the power of negative begin display style 1 third end style end exponent space equals space straight y to the power of begin display style 1 third end style end exponent over straight x to the power of begin display style 1 third end style end exponent space equals space minus open parentheses straight y over straight x close parentheses to the power of 1 third end exponent$
At (1, 1), slope of tangent $equals negative open parentheses 1 over 1 close parentheses to the power of 1 third end exponent space equals space minus 1$

$therefore space space space space space space equation space of space tangent space at space left parenthesis 1 comma space 1 right parenthesis space is space space space space space space space space space space space space straight y space minus 1 space equals space minus left parenthesis straight x minus 1 right parenthesis comma space space space or space space space space straight y minus 1 space equals space minus straight x plus 1 space space space or space space straight x plus straight y minus 2 space equals space 0 At space left parenthesis 1 comma space 1 right parenthesis comma space slope space of space normal space space equals space minus fraction numerator 1 over denominator negative 1 end fraction space equals space 1 therefore space space equaiton space of space normal space at space left parenthesis 1 comma space 1 right parenthesis space is space space space space space space space space space space space space space space space space space straight y space minus 1 space space equals space 1 space left parenthesis straight x minus 1 right parenthesis space space space space space or space space space space space straight y space minus 1 space equals space straight x minus 1 space space space or space space straight x minus straight y space equals space 0$