How do you find the equations of the tangent lines to the curve #y^2 - xy+10 = 0# at the points #(- 7, - 2)# and #(- 7, - 5)#?

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Answer 1 · 2017-06-23T01:24:33

For #(-7,-2)#: #" "y+2=-2/3(x+7)#

For #(-7,-5)#: #" "y+5=5/3(x+7)#

#d/dx(y^2-xy+10)=d/dx(0)#

#2y * dy/dx - (y+x * dy/dx) = 0#

#2y*dy/dx - x*dy/dx -y = 0#

#(2y-x)*dy/dx = y#

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

So now we have an expression for dy/dx. To find the slope at each point, we just plug in the X and Y values for those points.

The slope at #(-7,-2)# is:

#y/(2y-x)=(-2)/(2(-2)-(-7))#

#= (-2)/(-4+7) = -2/3#

So the equation of the tangent line is

#y+2=-2/3(x+7)#

The slope at #(-7,-5)# is:

#y/(2y-x) = (-5)/(2(-5)-(-7))#

#= (-5)/(-10+7) = 5/3#

So the equation of the tangent line is:

#y+5=5/3(x+7)#

Final Answer