Find a formula for the slope of the line tangent to an ellipse whose equation is of the form:

Question

x²/a² + y²/b² = 1

(Here a and b are nonzero constants) Then find the coordinate(s) of the point(s) where the tangent line is vertical (in terms of a and/or b)

Answer 1

If x²/a²+y²/b²=1 then we know right off the bat that the ellipse is centered about the origin with y-intercepts at y=±b and x-intercepts at x=±a.

The coordinates where the tangent lines are vertical are simply (a,0) and (-a,0).

Now using implicit differentiation,

2x/a²+(2y/b²)(dy/dx)=0

dy/dx =m=-(x/y)/(a/b)²

And we know from the original ellipse formula that y = ±b[1-(x/a)²]^1/2. So substituting,

dy/dx=slope=-(x/{±b[1-(x/a)²}^1/2)/(a/b)²

I'll leave it up to you to simplify.

Answer 2

Use implicit differentiation to find dy/dx:

d/dx( x²/a² ) + d/dx( y²/b² ) = d/dx(1)

(2x / a² ) + (2y / b² ) * dy/dx = 0

dy/dx = -(2x / a² ) / (2y / b² ) = -b2/a² * x/y

Since the ellipse is centered at (0,0), you can find the vertical tangent point simply by setting y = 0 in the original equation. This gives us x² = a² , and therefore x = ± a

CHECK: It is always good to check things like this by plotting in an online utility such as DESMOS:

https://www.desmos.com/calculator

Answer 3

dy/dx = (-bx/a)/sqrt(a² - x²⁾

^{the slope is vertical where x=±a and b=0}