ellipse problem

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:
$\sqrt{(x - 1)^2 + (y - 2)^2} + \sqrt{(x - 5)^2 + (y - 2)^2} = 16$

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

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

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

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

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

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

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

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

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

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

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

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

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

-Dan