# Implicit function, tangent line, HARDER THAN USUAL

• Oct 4th 2009, 10:20 PM
Tclack
Implicit function, tangent line, HARDER THAN USUAL
Ok, I'm sure all of you senior guys are tired of these problems, but this one is tricky. I'll stump you, I know it....if not for a few seconds:

The function

x^2 -4x +y^2 +3 = 0

find the equation of the 2 tangent lines in this function that pass through the origin.

My attempt:
As usual I would take the derivative

2x - 4 +2y dy/dx = 0
dy/dx= (2-x)/y

And NORMALLY I would plug in the coordinates of the other point (0,0) to get the slope "m" and put it into y=mx+b form. But division by 0 leads to some problems.

so I took the original equation and re-arranged for y, which got me:

y= sqrt[ 4x-x^2 -3]

and took the slope the old fashioned way

m=(y-0)/(x-0) which equals the derivative (2-x)/y
y/x = (2-x)/y
sqrt[ 4x-x^2 -3]/x = (2-x)/sqrt[ 4x-x^2 -3]

simplifying to:
4x-x^2 -3 = x (2-x)

x=3/2
but this is not the answer. The actual answer is +/- (sqrt3)/3

I know I should keep the x^2 because the answer is obviously a simplification after the quadratic formula has been used. Where did I go wrong?
• Oct 4th 2009, 11:54 PM
BobP
Forget the derivative, you don't need it.
Suppose that the equation of the tangent line is $y=mx.$
Substitute this into your given equation and solve for $x.$
If the line is to be a tangent this has to give you equal roots, (so $b^2-4ac=0).$
It will also give you two values of $m.$
Incidentally, to be on the safe side, you should also sketch the graph of your original function, (it's a circle centre (2,0) radius 1), to confirm that your results make sense.
• Oct 5th 2009, 07:08 AM
HallsofIvy
Or, just note that the graph of this is a circle with center at (2, 0) with radius 1. The two tangent lines from the origin form right triangles with one leg of length 1 (a radius of the circle) and hypotenuse of length 2 so the other leg has length $\sqrt{4- 1}= \sqrt{3}$. That is, the two points at which the tangent lines through the origin touch the circle are also on the circle with center at the origin and radius $\sqrt{3}$. Thus, (x,y) satisfy both $x^2- 4x+ y^2= -3$ and $x^2+ y^2= 3$. Subtracting the first equation from the second, -4x= -6 or x= 3/2. Of course, then $y^2= 3- 9/4= 3/4.$. The tangent lines are $y= \sqrt{3} x$ and $y= -\sqrt{3} x$.

Not quite as hard as you thought it was?

(BobP's method of finding the slope was Fermat's method from before Newton and Leibniz developed the calculus.)
• Oct 5th 2009, 07:21 AM
Tclack
Quote:

Originally Posted by BobP
Forget the derivative, you don't need it.
Suppose that the equation of the tangent line is $y=mx.$
Substitute this into your given equation and solve for $x.$
If the line is to be a tangent this has to give you equal roots, (so $b^2-4ac=0).$
It will also give you two values of $m.$
Incidentally, to be on the safe side, you should also sketch the graph of your original function, (it's a circle centre (2,0) radius 1), to confirm that your results make sense.

When I simplified, I had no a term, so putting my data into $b^2-4ac=0$
I get $2^2-4(0)c = 0$, so 4=0 doesn't make sense, maybe I'm misinterpreting.
• Oct 5th 2009, 07:26 AM
Tclack
What is the significance of 3/2 ?

Quote:

Or, just note that the graph of this is a circle with center at (2, 0) with radius 1.
And how were you able to determine the circles shape? That extra "4x" term really threw me off.

If you had to solve it without graphing, how would you do it?
• Oct 5th 2009, 08:32 AM
BobP
$x^2-4x+y^2+3=0.$

$y=mx.$

$x^2-4x+m^2x^2+3=0.$

$(1+m^2)x^2-4x+3=0.$

$b^2-4ac=16-12(1+m^2).$

You should find that
$y=\pm\frac{\sqrt{3}}{3}x.$

To show that the original equation is that of a circle, complete the square on $x.$
• Oct 5th 2009, 09:16 AM
Tclack
Ok, so I have a visual, attached at the bottom. I can't quite see how to solve this graphically with the pythagorean theorem. What's forming the right triangle?
• Oct 5th 2009, 10:49 AM
earboth
Quote:

Originally Posted by Tclack
Ok, so I have a visual, attached at the bottom. I can't quite see how to solve this graphically with the pythagorean theorem. What's forming the right triangle?

I've modified your sketch a little bit.

Keep in mind that the slope of a straight line is the tangens of the angle between the line and the x-axis.