How to find tangent line on the right spheroid with a parameter $a = 2.5$ and at point ($1,0.64565$)?

Question

Let $y^2=\dfrac{x^2(a-x)}{(a+x)}$ be given.

I am not quite sure how the right spheroid works in general. I have calculated the point on the curve by plugging the random value of $x = 1$ and got $y$ approximately $0.64565$, hence the title.

I am not quite sure how to calculate the tangent and normal line of the right spheroid, and what is the actual formula for the right spheroid in general?

Thanks in advance!

Answer 1

I have decided to make my comment an answer.

The operations involved should be in your skill set.

You must find the implicit derivative of with respect to $x$ using the equation of the curve and substitute the values of $a,x,y$ at the point of tangency to find the slope of the tangent line. Then you have the coordinates of a point on the tangent line and its slope, which is all you need to find the equation. The slope of the normal line is the negative reciprocal of the slope of the tangent line, so you can also find the equation of the normal line.

Note that there are two points on the curve with $x$-coordinate $1$.