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Application Of Derivatives

Question
CBSEENMA12035212

Find out the points on the curve straight x squared over 9 minus straight y squared over 16 space equals space 1 at which the tangents are (i) parallel to the x-axis and (ii) parallel to the y-axis.

Solution

 The equation of curve is  straight x squared over 9 minus straight y squared over 16 space equals space 1
Differentiating both sides w.r.t.x, we get,
                           fraction numerator 2 straight x over denominator 9 end fraction minus fraction numerator 2 straight y over denominator 16 end fraction dy over dx space equals space 0 space space space space space space or space space space space space space straight y over 8 dy over dx space equals space fraction numerator 2 straight x over denominator 9 end fraction
therefore space space space space space space space dy over dx space equals space fraction numerator 16 space straight x over denominator 9 space straight y end fraction
(i) For the points on the curve, where tangents are parallel to x-axis.
                        dy over dx space equals space 0 space space space space space rightwards double arrow space space space fraction numerator 16 straight x over denominator 9 straight y end fraction space equals space 0 space space space space rightwards double arrow space space space space straight x space equals space 0
Putting x = 0 in (1), we get,
                     0 minus straight y squared over 16 space equals space 1 space space space space space rightwards double arrow space space space space straight y squared space equals space minus 16 space space space space rightwards double arrow space space space space straight y space equals space plus-or-minus 4 space straight i
therefore  there is no real point at which the tangent is parallel to x-axis.
(ii) For the points on the curve, where tangents are parallel to y-axis.
                        dx over dy space equals space 0 space space space space space rightwards double arrow space space space space space fraction numerator 9 straight y over denominator 16 straight x end fraction space equals space 0 space space space rightwards double arrow space space space straight y space equals space 0
Putting y = 0 in (1), we get,
                     straight x squared over 9 minus 0 space equals space 1 space space space space space space space rightwards double arrow space space space space space space straight x squared space equals space 9 space space space space space space rightwards double arrow space space space space straight x space equals negative 3 comma space space 3
therefore space space space space points space are space left parenthesis negative 3 comma space 0 right parenthesis comma space space left parenthesis 3 comma space 0 right parenthesis

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