# Tangent to an ellipse

• Apr 23rd 2010, 04:29 AM
TsAmE
Tangent to an ellipse
Show by implicit differentiation that the tangent to the ellipse:

x^(2) / a^(2) + y^(2) / b^(2) = 1

at the point (x0, y0) is:

[ (x0x) / a^(2) ] + [ y0y / b^(2) ]

I got my gradient to be y' = -b^(2)x0 / a^(2)y0 but my tangent line was:

y = [ -b^(2)x0 / a^(2)y0 ] + [b^(2)x0^(2) / a^(2)y0^(2)]

Im not sure what the problem may be as I simply subbed in the gradient and (x0,y0) points into the y = mx + c equation
• Apr 23rd 2010, 05:21 AM
tonio
While trying to answer the following I transcribe part of the questions into LaTeX for clearity's sake

Quote:

Originally Posted by TsAmE
Show by implicit differentiation that the tangent to the ellipse:

x^(2) / a^(2) + y^(2) / b^(2) = 1

*** $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

at the point (x0, y0) is:

[ (x0x) / a^(2) ] + [ y0y / b^(2) ]

*** $\frac{x_0x}{a^2}+\frac{y_0y}{b^2}=1$ . Please do note the correction "=1" done to the above

I got my gradient to be y' = -b^(2)x0 / a^(2)y0 but my tangent line was:

y = [ -b^(2)x0 / a^(2)y0 ] + [b^(2)x0^(2) / a^(2)y0^(2)]

*** $y=\frac{-b^2x_0}{a^2y_0}+\frac{b^2x_0^2}{a^2y_0^2}$

We agree: $y'=\frac{dy}{dx}=-\frac{b^2x}{a^2y}$ and at the point $(x_0, y_0)$ we get what you wrote, but what you say is the tangent line's eq. cannot be

correct since, for example, that has no $x-$term in it...! The eq. of the tangent line to the ellipse at this point is in fact:

$y-y_0=-\frac{b^2x_0}{a^2y_0}(x-x_0)\Longrightarrow a^2y_0y=-b^2x_0(x-x_0)+a^2y_0$ $\Longrightarrow a^2y_0y+b^2x_0x=b^2x_0^2+a^2y_0^2$ , and now just

divide both sides by $a^2b^2$ and remember that $(x_0,y_0)$ is a point on the ellipse...and you'll get what you were asked!(Wink)

Tonio

Im not sure what the problem may be as I simply subbed in the gradient and (x0,y0) points into the y = mx + c equation

.
• Apr 23rd 2010, 08:48 AM
TsAmE
So you cant work it out with the y = mx + c equation? I carried on from your solution and got (y0y / b^2) + (x0x / a^2) = (x0 / a^2) + (y0 / b^2) but the right hand side doesnt equal 1?
• Apr 23rd 2010, 11:55 AM
HallsofIvy
Of course you can use "y= mx+ c". As Tonio said, y', at $(x_0, y_0)$ is equal to $-\frac{b^2x_0}{a^2y_0}$ so an equation with that slope passing through [tex](x_0, y_0) would be $y= -\frac{b^2x_0}{a^2y_0}(x- x_0)+ y_0$

Multiply that out and you will get "y= mx+ c".
• Apr 23rd 2010, 03:57 PM
TsAmE
Yeah but if I use y = mx + c and sub in the gradient and the x0 and y0 then the "x" and "y" essentially disappear, but the tangent line is suppose to contain an x and y
• Apr 24th 2010, 02:23 AM
tonio
Quote:

Originally Posted by TsAmE
Yeah but if I use y = mx + c and sub in the gradient and the x0 and y0 then the "x" and "y" essentially disappear, but the tangent line is suppose to contain an x and y

Uuh?? What do you mean by "the tangent line is supposed to contain x and y"?? What is not clear in the post I sent you?? (Wondering)

Tonio
• Apr 24th 2010, 04:00 AM
HallsofIvy
Quote:

Originally Posted by TsAmE
Yeah but if I use y = mx + c and sub in the gradient and the x0 and y0 then the "x" and "y" essentially disappear, but the tangent line is suppose to contain an x and y

I have no idea what you mean by this. "Sub in x0 and y0" where? Not for x and y- x0 and y0 should appear only in m and c.