Question

Find the points on the curve $\frac{x^2}{4}+\frac{y^2}{25}=1$ at which the tangents are parallel to x-axis

Answer 1

If the tangents are parallel to the x axis then the slope of the tangents are 0.
Differentiating the equation of the ellipse with respect to x gives us
$\frac{x}{2}+\frac{2y.y^{\prime }}{25}=0$.
Substituting $y^{\prime }=0$ since they are parallel to x axis,
$\frac{x}{2}+\frac{2y\left(0\right)}{25}=0$
Or
$x=0$
Therefore
$(\frac{x^2}{4}+\frac{y^2}{25}=1{)}_{x=0}$
$y^2=25$ or $y=\pm 5$.
Hence the points are $(x,y)=(0,5)$ and $(x,y)=(0,-5)$.