-->

Application Of Derivatives

Question
CBSEENMA12035214

Find points on the curve straight x squared over 4 plus straight y squared over 25 space equals space 1 at which the tangents are (i) parallel to the x-axis (ii) parallel to the y-axis.

Solution

The equation of curve is straight x squared over 4 plus straight y squared over 25 space equals space 1                  ...(1)
Differentiating both sides w.r.t. x, we get,
                      fraction numerator 2 straight x over denominator 4 end fraction plus fraction numerator 2 straight y over denominator 25 end fraction dy over dx space equals space 0 comma space space space space or space space space space space fraction numerator 2 straight y over denominator 25 end fraction dy over dx space equals space minus straight x over 2
therefore space space space space space dy over dx space equals space minus 25 over 4 straight x over straight y
(i) For the points on the curve, where tangent are parallel to x-axis,
                           dy over dx space equals space 0 space space space space rightwards double arrow space space space space space minus 25 over 4 straight x over straight y space equals space 0 space 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 plus straight y squared over 25 space equals space 1 space space space space space space rightwards double arrow space space space space space straight y squared space equals space 25 space space space space space rightwards double arrow space space space space space straight y space equals space minus 5 comma space 5
therefore space space space space space points space are space left parenthesis 0 comma space minus 5 right parenthesis comma space space left parenthesis 0 comma space 5 right parenthesis.
(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 minus fraction numerator 4 straight y over denominator 25 straight x end fraction space equals space 0 space space space space space space rightwards double arrow space space space space straight y space equals space 0
Putting y = 0 in (1), we get,
                                     straight x squared over 4 plus 0 space equals space 1 space space space space rightwards double arrow space space space straight x squared space equals space 4 space space space space rightwards double arrow space space space space straight x space equals space minus 2 comma space 2
therefore space space space space points space are space left parenthesis negative 2 comma space 0 right parenthesis comma space space left parenthesis 2 comma space 0 right parenthesis.