Find the equations of the lines tangent to
x^2+y^2-x+3y-2=0 with the direction of
the line 2x+2y-5=0.

Question

Find the equations of the lines tangent to
x^2+y^2-x+3y-2=0 with the direction of
the line 2x+2y-5=0.

amistre64 · Answer

define y' implicitly, and equate it to the slope of the given line

owlcoffee · Answer

You mean to first put it in ordinary form to deduce it's slope?

amistre64 · Answer

the line part,  sure.  that is one way to find its slope.

amistre64 · Answer

an implicit rundown for:  

x^2+y^2-x+3y-2=0

2x + 2y y' -1+ 3y' = 0

solve for y' and equate it with the slope of the line

owlcoffee · Answer

by Equating, do you mean evaluating?

amistre64 · Answer

hmm, maybe.  depends on where youre from i spose.

the slope of the given line is -1, right?

and y' defines the slope of the line at any given point on the curve.

when y' = -1, we know the point at which the line is tangent to the curve; and can use the pont slope form of a line to define the tangent line equations with

amistre64 · Answer

2x + 2y y' -1+ 3y' = 0

y' (2y + 3) = 1-2x

y'  = (1-2x)/(2y+3)

(1-2x)/(2y+3) = -1

1 - 2x = -2y - 3

amistre64 · Answer

using y = x-2 as a solution; sub it into the curve

x^2+y^2-x+3y-2=0

x^2+(x-2)^2-x+3(x-2)-2=0; and solve for x (and y as a result).

amistre64 · Answer

i have class starting so I have to leave ... good luck

owlcoffee · Answer

Thank you for your time, I am very grateful.
Good luck with your class ~

amistre64 · Answer

in solving for x

x^2+(x-2)^2-x+3(x-2)-2=0

x^2+(x^2-4x+4)-x+(3x-6)-2=0

2x^2 - 2x - 4=0

2 (x^2 - x - 2)=0

we see that x = -1, or 2
giving us:  y = -1-2, or 2-2

to construct the tangent lines with a slope of -1

owlcoffee · Answer

I see.
I'll try something, hhow about using the line family that go through that point, and consider the lines tangent to the circle.
All will be paralell so ther shouldn't be any trouble.

amistre64 · Answer

since:  x^2+y^2-x+3y-2=0 , is a circle, you could perp the slope of the given line to 1, and attach it to the center of the circle and determine where it crosses the circle at as well.  But that is more of a geometric interpretation than a calculus based approach :)

owlcoffee · Answer

basically the tangent lines and the line that crosses the center should have the same slope.
using the line family would give the same slope as if using implicit differentiation.
Again, all those are methods hehe...

amistre64 · Answer

indeed :)