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

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

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Parallel to y-axis means the slope of the tangent is  \infty , means the slope of normal is 0
We know that slope of the 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} = 144(1-32x)
\frac{dy}{dx} = \frac{-32x}{18y} = \infty
Slope of normal = -\frac{dx}{dy} = \frac{18y}{32x} = 0
From this we can say that y = 0
Now. when y = 0,  \frac{x^2 }{9} + \frac{0^2 }{16} \Rightarrow 1 = x = \pm 3
Hence, the coordinates are (3,0) and (-3,0)

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