The points on the curve (x^2)/16+(y^2)/25=1 at which tangents are par…

$\frac{x^2}{16}+\frac{y^2}{25}=1$         .......(1)

Differentiating (1)  w.r.t  x

$\frac{2 x}{16}+\frac{2 y}{25} \frac{d y}{d x}=0$

as tangent is parallel to x-axis

$\frac{dy}{dx} = 0$

so  $\frac{2 x}{16}+0=0$

$\Rightarrow x=0$

from (1)

putting x= 0

we get $\frac{y^2}{25}=1 \Rightarrow y^2 = 25$

So $y= \pm 5$

at points (0, ±5) tangents are parallel to x-axis.