Find points on the curve `(x^2)/9+(y^2)/(16)=1` at which the tangents are
(i) parallel to x-axis
(ii) parallel to y-axis.

Explanation

The equation of the given curve is `x^2/9+y^2/16=1`

On differentiating both sides with respect to `x`, we have:

`(2x)/9+((2y)/16)dy/dx=0impliesdy/dx=(-16x)/(9y)`

(i)
The tangent is parallel to the x-axis if the slope of the tangent is `0`.
i.e. `(-16x)/(9y)=0` which is possible if `x=0`

then `x^2/9+y^2/16=1`

for `x=0impliesy^2=16implies y=+-4`

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