Question

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x² - 6xy + 12y² = 28, (2, 1) (ellipse)

Answer 1

To find the tangent line equation at (2, 1) for the given curve, implicit differentiation is used to find the derivative. The slope of the tangent line is found to be 1/6. The final equation of the tangent line is
`y = (1/6)x + 2/3`.

Step-by-step explanation:

Implicit Differentiation and Tangent Line Equation

To find the equation of the tangent line to the curve
`x² - 6xy + 12y² = 28` at the point (2, 1), we must first use implicit differentiation to find the slope of the tangent line at that point. Differentiating both sides of the equation with respect to x:

`2x - 6(y + xy' + 12(2yy') = 0`

Rearranging the terms to isolate
`y'`, we find:

`2x - 6y - 6xy' + 24yy' = 0`

Combine like terms:

`(-6x + 24y)y' = 6y - 2x`

Then, solve for
`y'`:

`y' = (6y - 2x) / (-6x + 24y)`

Substituting
`x = 2` and
`y = 1` into the derivative gives us the slope of the tangent line at the point (2, 1):

`m = y'(2, 1) = (6(1) - 2(2)) / (-6(2) + 24(1)) = (6 - 4) / (-12 + 24) = 2/12 = 1/6`

Now we have the slope, we can use the point-slope form to find the equation of the tangent line:

`y - 1 = (1/6)(x - 2)`

The final step is to convert this into slope-intercept form, which illustrates the equation of the tangent line:

`y = (1/6)x + 2/3`