I'm a dunce

Cataclysm · #1 $Old$ 11-15-2008, 08:20 PM

This question is giving me a headache, and it's for homework. I usually can do these kinds of questions, but I feel really unmotivated right now.. Someone can help me solve it so I can see again..

Find the equations of the focal chord that cuts the curve x^2=8y at (-4, 2)

Soroban · #2 $Old$ 11-15-2008, 09:04 PM

Hello, Cataclysm!

If you made a sketch, the answer is obvious . . .

Quote:

Find the equation of the focal chord that cuts the curve $x^2\:=\:8y$ at (-4, 2)

From the equation, $x^2 \:=\:8y$ , we know all this:
. . The parabola opens upward.
. . Its vertex is at the origin.
. . The focus is at (0, 2).

Code:

                |
     *          |           *
                |
      *        F|(0,2)     *
 (-4,2)*- - - - o - - - -*
         *      |      *
            *   |   *
    ------------*------------
                |

The focal chord that contains (-4, 2) is the horizonal line: . $y \:=\:2$

Cataclysm · #3 $Old$ 11-15-2008, 11:54 PM

Yeah, that's what I thought it was, but I just couldn't explain myself properly. Thanks very much!