# Curve's point A tangent line passes through a point P. What is the distance of AP

• May 15th 2011, 08:11 AM
gundanium
Curve's point A tangent line passes through a point P. What is the distance of AP
Let A be a point on the curve C: $\displaystyle {x}^{ 2} + {y}^{ 2} - 2x + 4 = 0$. If the tangent line at A passes through P(4,3), then the length of AP is?

So i'm stuck midway at the solution for this one.
So far I've gotten these equations
$\displaystyle {y}^{ '} = \frac{1-x}{ y} = m$ ; the tangent line slope at any point on the curve (by implicit differentiation)
$\displaystyle m = \frac{y-3}{x-4}$; the slope at point P; acquired by y-k=m(x-h)

and when I put those two equations together I get:
$\displaystyle {-x}^{ 2} + 5x - 4 + 3y= {y}^{ 2}$

And when I input this equation back into the original curve i end up with
x= -y

I'm stuck at this point because when I plug in the last equation back to original curve again, I'm getting an complex solution.

Oh Gosh. I am sorry. That's supposed to be
$\displaystyle {x}^{ 2} + {y}^{ 2} - 2x - 4 = 0$
• May 15th 2011, 09:16 AM
Jester
You sure you got the curve C right. Completing the square in $\displaystyle x$ gives

$\displaystyle (x-1)^2 + y^2 + 3 = 0$

There are no real solutions to this equation.
• May 15th 2011, 09:25 AM
gundanium
Quote:

Originally Posted by Danny
You sure you got the curve C right. Completing the square in $\displaystyle x$ gives

$\displaystyle (x-1)^2 + y^2 + 3 = 0$

There are no real solutions to this equation.

Oh Gosh. I am sorry. That's supposed to be
{x}^{ 2} + {y}^{ 2} - 2x - 4 = 0
• May 15th 2011, 09:39 AM
earboth
Quote:

Originally Posted by gundanium
Let A be a point on the curve C: $\displaystyle {x}^{ 2} + {y}^{ 2} - 2x - 4 = 0$. If the tangent line at A passes through P(4,3), then the length of AP is?

...

1. Your curve is a circle with the center M(1, 0) and the radius $\displaystyle r = \sqrt{5}$.

2. The triangle MPA is a right triangle with MP as the hypotenuse and MA = r.

3. Use Pythagorean theorem to determine the length of AP.
Spoiler:
$\displaystyle (MP)^2=(AP)^2+r^2~\implies~|\overline{AP}|=\sqrt{1 3}$
• May 15th 2011, 09:42 AM
Jester
This is the way I would do it. Suppose the point on the circle is $\displaystyle (a,b).$ Then $\displaystyle (a,b)$ satisfies

$\displaystyle a^2+b^2-2a-4 = 0\;\;\;(1)$.

The tangent to the curve at this point is

$\displaystyle m = \dfrac{1-a}{b}$

and the slope of the line connecting $\displaystyle (a,b)$ to $\displaystyle (3,4)$ is

$\displaystyle m = \dfrac{b-3}{a-4}$

These must equal, hence

$\displaystyle \dfrac{b-3}{a-4} = \dfrac{1-a}{b}$

or

$\displaystyle a^2+b^2 - 5a - 3b = -4\;\;\;(2)$

You are required to find the distance between the two points or

$\displaystyle d = \sqrt{(a-4)^2+(b-3)^2}$
$\displaystyle = \sqrt{a^2+b^2 - 8a -6b +25}$

Now $\displaystyle 2\times(2)-(1)$ gives

$\displaystyle a^2+b^2 - 8a - 6b + 12 = 0$

so

$\displaystyle d = \sqrt{25-12} = \sqrt{13}$

Edit: Earthboth's solution is so much more elegant (and of course easier!)
• May 15th 2011, 09:49 AM
gundanium
Quote:

Originally Posted by Danny
Now $\displaystyle 2\times(2)-(1)$ gives

Why? How did you come up with this?
• May 16th 2011, 04:57 AM
Jester
This is what I have

$\displaystyle a^2+b^2-2a = 4\;\;\;(1)$

$\displaystyle a^2+b^2 - 5a - 3b = -4\;\;\;(2)$

this is what I want

$\displaystyle a^2+b^2 -8a - 6b\;\;\;(3)$

Multiply $\displaystyle p \times (1) + q \times (2) = (3)$ (LHS only) and equate like terms. As it turns out, there is a unique $\displaystyle p$ and $\displaystyle q$ solution.