The following equation defines a family of plane curves. x3 + y3 - 3axy = c wher

Question

The following equation defines a family of plane curves. x3 + y3 - 3axy = c where c ( - infinity; + infinity). A unique family member passes through every point in the XY-plane except (0. 0) and (a, a). Let (alpha, beta) be a point on one of these plane curves. Your particular parameter and coordinates are a = -8 and (alpha, beta) = (-4,-8). Find an explicit equation for the tangent at (beta, alpha) = (-8,-4). Find the exact point of intersection of the tangent lines through points (alpha, beta) =( -4,-8 ).and (beta, alpha) = (-8, -4) Using MATLAB's ezplot, or similar software that does implicit plots, add to your plot in (b), the tangent lines you have found in (e) (ii) and (f) (iii).

Explanation / Answer

From the implicit differentiation:

3x^2 + 3y'y^2 - 3ay - 3axy' = 0 -> y' = (ay - x^2)/(y^2 - ax) = (-8y - x^2)/(y^2 + 8x)

at (-8,-4):

y' = (32 - 64)/(16 - 64) = 2/3

So the tangent line equation is:

y - (-4) = 2/3 (x - (-8))

or

y = 2/3 x + 4/3

2)

at (-4,-8) :

y' = (64 - 16)/(64 - 32) = 3/2

So the tangent line equation is:

y - (-8) = 3/2 (x - (-4))

or

y = 3/2 x - 2

To find the point of intersection we have:

y = 2/3 x + 4/3

and

y = 3/2 x - 2

Therefore:

2/3 x + 4/3 = 3/2 x - 2 -> 5/6 x = 10/3 -> x = 4 -> y = 4

So the point of intersection is (4,4)

3)

Use the following code:

ezplot('x^3+y^3+24xy=192')

ezplot('3y-2x-4')

ezplot('2y-3x+4')

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