# 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

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)

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