The equation 5x2 - 6xy + 5y2 = 16 represents an ellipse. A) Find dy/dx at (1, -1)?

Question

The equation 5x2 - 6xy + 5y2 = 16 represents an ellipse.

A) Find dy/dx at (1, -1)

B) Find two points on the ellipse at which the tangent is horizontal.

Please show work. Thanks

Answer 1

5x^2 -6xy+5Y^2 -16 = 0
10x dx-6x dy-6y dx +10 y dy = 0
(10 x -6y) dx = (6x -10 Y) dy
dy/dx = (10 x - 6 Y)/(6x-10 Y)

at (1,-1), dy/dx = (10+6)/(6+10) = 1 Ans A.

B) dy/dx = 0: 10 x = 6 Y: x = .6 Y

5* (.6y)^2 - 6 * y * (.6Y) + 5 * y^2 =
1.8Y^2 -3.6Y^2+5y^2 = 3.2 y^2 = 16
3.2 Y^2 = 16
y = +sqrt 5, - sqrt 5
X = +.6*sqrt 5, - .6 sqrt 5

Answer 2

Use implicit differentiation:
5x^2 - 6xy + 5y^2 = 16
=> 10x - 6x dy/dx - 6y + 10y dy/dx = 0
=> (5y - 3x) dy/dx = 3y - 5x
=> dy/dx = (3y - 5x) / (5y - 3x)

A) At (1, -1), dy/dx = (-3 - 5) / (-5 - 3) = 1.
B) dy/dx = 0 => 3y = 5x
5x^2 - 2x(3y) + 5/9 (3y)^2 = 16
=> 5x^2 - 2x(5x) + 5/9 (5x)^2 = 16
=> x^2 (5 - 10 + 125/9) = 16
=> x^2 = 9(16) / 80
=> x = ±(3/√5)
y = 5x/3 = ±(√5)
So the two points are (3/√5, √5) and (-3/√5, -√5).

Answer 3

you have to differentiate implicitly:

10x - 6y - 6x*dy/dx + 10y*dy/dx = 0
10x-6y = dy/dx(6x-10y)
dy/dx = (10x-6y)/(6x-10y)
divide each term by 2:
dy/dx = (5x-3y)/(3x-5y)

dy/dx at (1,-1) = (5*1 - 3*-1)/(3*1 - 5*-1) = 1

The tangent is horizontal when dy/dx = 0, which would be when the numerator equals 0. So 5x-3y = 0 --> y = 5x/3. We need to subsitute that back into the original equation of the ellipse and solve for x:

5x² - 6x(5x/3) +5(5x/3)² = 16
5x² - 10x² + 125x²/9 = 16
45x² - 90x² + 125x² = 144
80x² = 144
x² = 9/5
x = ±3/√5 = ±3√5 /5

Since y = 5x/3, then the corresponding y-values are ±√5.