ellipse problem

• Feb 1st 2007, 07:07 AM
ellipse problem
A point moves in the coordinate plane so that the sum of its distances from (1,2) and (5,2) is equal to 16. Find the equation of the path of the moving point...

:confused:
• Feb 1st 2007, 08:43 AM
topsquark
Quote:

Originally Posted by ^_^Engineer_Adam^_^
A point moves in the coordinate plane so that the sum of its distances from (1,2) and (5,2) is equal to 16. Find the equation of the path of the moving point...

:confused:

This is the classic problem to derive the equation of an ellipse with foci at (1, 2) and (5, 2).

Call a point on the curve (x, y). Then the distance formula says:
$\displaystyle \sqrt{(x - 1)^2 + (y - 2)^2} + \sqrt{(x - 5)^2 + (y - 2)^2} = 16$

Get rid of the radicals:
$\displaystyle \sqrt{x^2 + y^2 - 2x - 4y + 5} + \sqrt{x^2 + y^2 - 10x - 4y + 29} = 16$

$\displaystyle \sqrt{x^2 + y^2 - 2x - 4y + 5} = 16 - \sqrt{x^2 + y^2 - 10x - 4y + 29}$

$\displaystyle x^2 + y^2 - 2x - 4y + 5 = (16 - \sqrt{x^2 + y^2 - 10x - 4y + 29})^2$

$\displaystyle x^2 + y^2 - 2x - 4y + 5 = 256 -$$\displaystyle 32 \sqrt{x^2 + y^2 - 10x - 4y + 29} + (x^2 + y^2 - 10x - 4y + 29)$

$\displaystyle 32 \sqrt{x^2 + y^2 - 10x - 4y + 29} = 280 - 8x = 8(35 - x)$

$\displaystyle 4 \sqrt{x^2 + y^2 - 10x - 4y + 29} = 35 - x$

$\displaystyle 16 (x^2 + y^2 - 10x - 4y + 29) = (35 - x)^2$

$\displaystyle 16x^2 + 16y^2 - 160x - 64y + 464 = 1225 - 70x + x^2$

$\displaystyle 15x^2 - 90x + 16y^2 - 64y = 761$

Now complete the square on x and y:
$\displaystyle 15(x^2 - 6x) + 16(y^2 - 4y) = 761$

$\displaystyle 15(x^2 - 6x + 9) + 16(y^2 - 4y + 4) = 761 + 15 \cdot 9 + 16 \cdot 4$

$\displaystyle 15(x - 3)^2 + 16(y - 2)^2 = 960$

Divide both sides by 960:
$\displaystyle \frac{(x - 3)^2}{64} + \frac{(y - 2)^2}{60} = 1$

-Dan