This one's bothering me..
Find equations for two lines through the origin that are tangent to the curve
MIT 18.01 Single Variable Calculus (OCW)
Stacey Warren - Expert brainly.com
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a picture is usually helpful.
complete the squares to determine that its a circle centered at (2,0) with a radius of 1
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therefore C(2,0), Radius=1
Tangent lines should be in the form of:
Gotta find 'a'!!
we get from the drawing that \[\sin a = 1/2\]
bearing in mind that 'a' is the slope of the tangent line.
My solution is: \[y=+-(\pi*x)/\]
your a little off i think
for starters; pi/4 = 45 degrees, not 30
and it is the tangent of 30 degrees that defines the slope; not the sine
the tangent lines are then:
y = tan(30)x and y = -tan(30)x
oh, guess i missed that! haha
you are right, that was the only solution I could think of, my skills are a litte rusty!
you had the right notion, just a little off in the presentation ;)
i guess it settles it then, as \[\tan 30º = \sqrt3/3\]
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.
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