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  • Find points on the curve x ^2 / 9 +y^2 / 16 = 1 at which the tangents are parallel to x-axis

13) Find points on the curve \frac{x^2 }{9} + \frac{y^2 }{16} = 1 at which the tangents are  (i) parallel to x-axis 

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Parallel to x-axis means slope of tangent is 0
  We know that slope of tangent at a given point on the given curve is given by  \frac{dy}{dx}
Given the equation of the  curve is 
\frac{x^2 }{9} + \frac{y^2 }{16} = 1 \Rightarrow 9y^2 = 144(1-16x^2)
18y\frac{dy}{dx} = -32x
\frac{dy}{dx} = \frac{(-32x)}{18y} = 0 \Rightarrow x = 0
From this, we can say that x = 0
Now. when x = 0  ,     \frac{0^2 }{9} + \frac{y^2 }{16} = 1\Rightarrow \frac{y^2}{16} = 1 \Rightarrow y = \pm 4
Hence, the coordinates are (0,4) and (0,-4)

 

Posted by

Gautam harsolia

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