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Short Answer Type

71. Find the equation of the tangent and normal to the curve x = a t² , y = 2 at at the point ‘t’.

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Long Answer Type

72. Find the equation of the tangent at

straight t space equals space straight pi over 4 space to space the space curve space straight x space equals sin space 3 straight t comma space space space space space straight y space equals space cos 2 straight t.

73 Views

Short Answer Type

73.

Find the equation of tangent to the curve given by x = a sin³ t, y = b cos³ t at a point where $straight t equals space straight pi over 2.$

102 Views

Long Answer Type

74.
Find the equation of tangent to the curve:
$straight x space equals space straight theta plus space sin space straight theta comma space space space straight y space equals space 1 plus space cos space straight theta space space at space straight theta equals space straight pi over 4$

The equations of curve are $straight x space equals straight theta space plus space sin space straight theta comma space space space space straight y space equals space 1 plus space cos space straight theta$
$At space straight theta space equals space straight pi over 4 comma space space space straight x space equals space straight pi over 4 plus sin space straight pi over 4 space equals space straight pi over 4 plus fraction numerator 1 over denominator square root of 2 end fraction comma space space space space straight y space equals 1 plus cos space straight pi over 4 space equals space 1 plus fraction numerator 1 over denominator square root of 2 end fraction therefore space space space space point space of space contact space is space open parentheses straight pi over 4 plus fraction numerator 1 over denominator square root of 2 end fraction comma space space 1 plus fraction numerator 1 over denominator square root of 2 end fraction space close parentheses. space space space space space space space space space space space space space space space dx over dθ space equals space 1 plus cos space straight theta. space dy over dθ space equals space minus sin space straight theta space space space space dy over dx space equals space fraction numerator begin display style dy over dθ end style over denominator begin display style dx over dθ end style end fraction space equals space minus fraction numerator sin space straight theta over denominator 1 plus cos space straight theta end fraction$
$At space space space straight theta space equals space straight pi over 4 comma space dy over dx space equals space minus fraction numerator sin space begin display style straight pi over 4 end style over denominator 1 plus cos space begin display style straight pi over 4 end style end fraction space equals space minus fraction numerator begin display style fraction numerator 1 over denominator square root of 2 end fraction end style over denominator 1 plus begin display style fraction numerator 1 over denominator square root of 2 end fraction end style end fraction space equals negative fraction numerator 1 over denominator square root of 2 plus 1 end fraction comma$ which is slope of tangent

$therefore space space space space space space equation space of space tangent space to space the space curve space at space straight theta space equals space straight pi over 4 space space is$

$straight y minus open parentheses 1 plus fraction numerator 1 over denominator square root of 2 end fraction close parentheses space equals space minus fraction numerator 1 over denominator square root of 2 plus 1 end fraction open square brackets straight x minus open parentheses straight pi over 4 plus fraction numerator 1 over denominator square root of 2 end fraction close parentheses close square brackets$

or $straight y minus open parentheses 1 plus fraction numerator 1 over denominator square root of 2 end fraction close parentheses space equals space minus left parenthesis square root of 2 minus 1 right parenthesis space open square brackets straight x minus open parentheses straight pi over 4 plus fraction numerator 1 over denominator square root of 2 end fraction close parentheses close square brackets$
$open square brackets because space space space space space fraction numerator 1 over denominator square root of 2 plus 1 end fraction space equals space fraction numerator 1 over denominator square root of 2 plus 1 end fraction cross times fraction numerator square root of 2 minus 1 over denominator square root of 2 minus 1 end fraction space equals space fraction numerator square root of 2 minus 1 over denominator 2 minus 1 end fraction space equals space square root of 2 minus 1 close square brackets$

73 Views

75.

Find the equation of tangent to the curve:
$straight x space equals space 1 minus cos space straight theta comma space space space straight y space equals space straight theta space plus space sin space straight theta space space at space straight theta space equals space straight pi over 4.$

74 Views

76.

Find the equation of tangent to the curve:
$straight x space equals space 1 plus cos space straight theta comma space space space space straight y equals space straight theta plus sinθ space space at space straight theta space equals space straight pi over 4.$

72 Views

77.

Find the equation of the tangent and normal to the curve
$straight x space equals space cost comma space space space space space straight y space equals space sin space straight t space space space at space straight t space equals space straight pi over 4$

78 Views

Short Answer Type

78.

Find the equation of the tangent to the curve
$straight y space equals space fraction numerator 1 plus sinx over denominator cosx end fraction space at space straight x space equals space straight pi over 4.$