CALC equations for tangent and normal lines!!?

Question

Find the eq for tan and normal lines at point (a,a)
x^3-axy+3ay^2 = 3a^3

Answer 1

x^3-axy+3ay^2 = 3a^3
take the derivative with respect to x

3x^2 -axy'- ay +6ay y' =0
then y'(-ax+6ay)

Answer 2

For some reason, lolitabellabella didn't complete her answer.
Here's her beginning:
x^3-axy+3ay^2 = 3a^3
take the derivative with respect to x

3x^2 -axy'- ay +6ay y' =0

(I deleted her last line, which was not complete.)

Here's the rest:
then y'( -ax + 6ay) = -3x^2 + ay
so y' = (-3x^2 + ay) / ( -ax + 6ay)

At the point (a,a), y' = ( -3a^2 + a^2) / ( -a^2 + 6a^2)

You should be able to cancel the a^2's in the numerator and denominator and reduce that to a simple number for y'. That's the slope.

The equation for the tangent line has that slope and passes through (a,a). You should be able to write the equation for the tangent line.

The normal is, of course, perpendicular to the tangent. You can find it's slope by taking the inverse of the tangent's slope and changing the sign. (Example: If tangent's slope is -2, then normal line's slope is -1/(-2) = 0.5)

So now you can calculate the slope of the normal line, and you know that the line passes through (a,a). You should be able to write the equation for the line with the desired slope that passes through that point.

Good luck!

Answer 3

So it type of sounds like you're off to a commence. a classic line is perpendicular to a tangent line. So locate the tangent slope first, and then use that to locate the conventional slope through dividing -a million through the tangent slope. Use the chain and ability regulations to locate the by-product of that function. y(x) = (x^2 - a million)^(a million/3) y'(x) = (a million/3)(2x)(x^2 - a million)^(-2/3) y'(3) = a million/2 it quite is the slope of the tangent line. The slope of the conventional line is -a million/(a million/2) = -2 Write an equation in the point slope sort to finish the challenge.