anonymous
  • anonymous
This one's bothering me.. Find equations for two lines through the origin that are tangent to the curve (x^2) -4x+(y^2)+3=0
MIT 18.01 Single Variable Calculus (OCW)
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
a picture is usually helpful. complete the squares to determine that its a circle centered at (2,0) with a radius of 1 |dw:1333203784659:dw|
amistre64
  • amistre64
as is; you have congruent triangles formed
amistre64
  • amistre64
|dw:1333205069749:dw|

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anonymous
  • anonymous
\[(x^2-4x+4)+(y^2-0)=1\] \[(x-2)^2+(y-0)^2=1\] therefore C(2,0), Radius=1 |dw:1333226481119:dw| Tangent lines should be in the form of: \[y=a*x\] \[y=(-a)*x\] Gotta find 'a'!! we get from the drawing that \[\sin a = 1/2\] \[a=30º=\pi/4\] bearing in mind that 'a' is the slope of the tangent line. My solution is: \[y=+-(\pi*x)/\]
anonymous
  • anonymous
\[y=(\pi*x)/4\] \[y'=(\pi*x)/4\]
amistre64
  • amistre64
your a little off i think
amistre64
  • amistre64
for starters; pi/4 = 45 degrees, not 30 and it is the tangent of 30 degrees that defines the slope; not the sine
amistre64
  • amistre64
the tangent lines are then: y = tan(30)x and y = -tan(30)x
anonymous
  • anonymous
oh, guess i missed that! haha you are right, that was the only solution I could think of, my skills are a litte rusty!
amistre64
  • amistre64
you had the right notion, just a little off in the presentation ;)
anonymous
  • anonymous
i guess it settles it then, as \[\tan 30º = \sqrt3/3\] \[y=+-(\sqrt3*x)/3\]
s3a
  • s3a
I'd say: 1) Implicit differentiation. 2) Isolate y'. 3) Make y' = 0. 4) Solve for those values of x. That's for the slopes. To get the y-intercept: Plug that same value of x you got into the y equation (the one before the differentiation) and find the y-intercept. This slope and y-intercept together should be each individual answer. I'm tired as I write this so if I'm wrong, someone say so.
amistre64
  • amistre64
the "make y' = 0" part is a bit askew. we want the slope of the tangent line to the circle to pass thru the origin. Making the slope 0 never really finds the lines we seek. if im reading your response correctly that is

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