Implicit Differentiation: x^2/a^2 + y^2/b^2 = 1

[SCGirl]

Hi, I just can't seem to get the right answer. The problem is:

Show by implicit differentiation that the tangent to the ellipse:

. . .x^2/a^2 + y^2/b^2 = 1

at the point (x1, y1) is:

. . .x1x/a^2 + y1y/b^2 = 1

I used implicit differentiation and got it to:

. . .dy/dx = a^2/2x + b^2/2y

I'm just stuck, please help.

 

[galactus]

By implicit diff, \(\displaystyle \L\\\frac{dy}{dx}=\frac{-b^{2}}{a^{2}}\cdot\frac{x_{1}}{y_{1}} \;\ \text{if} \;\ y_{1}\neq{0}\),

the tangent line is:

\(\displaystyle \L\\y-y_{1}=\frac{-b^{2}}{a^{2}}\cdot\frac{x_{1}}{y_{1}}(x-x_{1}),\\

a^{2}y_{1}y-a^{2}y_{1}^{2}=-b^{2}x_{1}x+b^{2}x_{1}^{2},\\

b^{2}x_{1}x+a^{2}y_{1}y=b^{2}x_{1}^{2}+a^{2}y_{1}^{2},\)

but \(\displaystyle \L\\(x_{1}, y_{1})\) is on the ellipse so \(\displaystyle \L\\b^{2}x_{1}^{2}+a^{2}y_{1}^{2}=a^{2}b^{2}\)

and the tangent line is:

\(\displaystyle \L\\b^{2}x_{1}x+a^{2}y_{1}y=a^{2}b^{2},\\
\frac{x_{1}x}{a^{2}}+\frac{y_{1}y}{b^{2}}=1.\)

If \(\displaystyle y_{1}=0\) then \(\displaystyle x_{1}=\pm{a}\) and the tangent lines are \(\displaystyle x=\pm{a}\) which also follow from \(\displaystyle \frac{x_{1}x}{a^{2}}+\frac{y_{1}y}{b^{2}}=1\)

Will that do?.

 

[SCGirl]

ok that makes sense. thank you!