How do you find the coordinates of the points on the curve #x^2-xy+y^2=9# where the tangent line is vertical?

Question

Evelyn Agard · Accepted Answer

Please see the sketch of a solution below.

Find #dy/dx# using implicit differentiation. #dy/dx = -(y-2x)/(x-2y)# The tangent will be vertical when #dy/dx# approaches #oo#, which happens at #y = 1/2x# Now substitute #1/2x# for #y# in the original equation and solve to get #x=+- 2sqrt3#. Finish by using #y = 1/2x# to get the #y# coordinates. So the points are #(2sqrt3, sqrt3)# and #(-2sqrt3, -sqrt3)#

Isaiah Allen · Answer

Tangents # x=+-2sqrt3#. Points of contact : #+-sqrt3( 1. 2, )#. See the Socratic tangents-inclusive graph of this ellipse.

x^2+y^2-xy-9=0. represents an ellipse. In the standard form, this is #(x+y)^2/36+(x-y)^2/12=1# Let us find #x'=(dx)/(dy)#. # 2x x'+2y-x'y-x=0,# giving for the vertical direction #2y=x#, when #x'=1/(y')=1/oo=0#. Substituting in the equation, #x^2+x^2/4-x^2/2=9#, giving #x = +-sqrt3#. The points of contact fo the tangents are #+-sqrt3( 1, 2 )#. So, the vertical tangents are #x=+-2sqrt3#. graph{((x+y)^2/36+(x-y)^2/18-1)(x-2sqrt3-.35+.01y)(x+2sqrt3+.35+.01y)=0 [-10, 10, -5, 5]}

Charles Anctil · Answer

undefined To find the coordinates of the points on the curve where the tangent line is vertical, we need to determine the points where the derivative of the curve with respect to x is undefined.

First, we differentiate the equation x^2 - xy + y^2 = 9 with respect to x using implicit differentiation:

d/dx (x^2 - xy + y^2) = d/dx (9)
2x - (x(dy/dx) + y) + 2y(dy/dx) = 0

Next, we solve for dy/dx:

dy/dx = (2x - y) / (x + 2y)

For the tangent line to be vertical, the derivative dy/dx must be undefined. This occurs when the denominator (x + 2y) equals zero.

Setting x + 2y = 0, we can solve for y:

y = -x/2

Substituting this value of y back into the original equation, we get:

x^2 - x(-x/2) + (-x/2)^2 = 9
x^2 + x^2/2 + x^2/4 = 9
(7x^2)/4 = 9
7x^2 = 36
x^2 = 36/7
x = ± √(36/7)

Using the equation y = -x/2, we can find the corresponding y-values:

For x = √(36/7):
y = -√(36/7)/2

For x = -√(36/7):
y = √(36/7)/2

Therefore, the coordinates of the points on the curve x^2 - xy + y^2 = 9 where the tangent line is vertical are:

(√(36/7), -√(36/7)/2) and (-√(36/7), √(36/7)/2)