Let P be the parabola in the plane determined by the equation y=x^2. Suppose a circle C intersects P at four distinct points. If three of these points are (-28,784),(-2,4)&(13,169) find the sum of the distances from the focus of P to all four of the intersection points.
Very confused through the wording of the question.
We need to solve this system .....it's a little tedious, so I used WolframAlpha to solve it....however....I could go through it if you wanted.....
(-28- h)^2 + (784 - k)^2 = r^2
(-2 - h)^2 + ( 4 - k)^2 = r^2
(13 - h)^2 + (169 - k)^2 = r^2
The center of the circle is at (h, k) = (-2475, 312) and r^2 = 6210593
The equation of the circle is ( x+ 2475)^2 + (y - 312)^2 = 6210593
Here's a graph : https://www.desmos.com/calculator/7uay3uslee
We can find the other "x" intersection point thusly ...sub x^2 for y in the equation of the circle and we have that
(x^2 + 2475)^2 + (x^2 - 312)^2 = 6210593
I also used WolramAlpha to solve this....again....I can go through it if you want
The last intersection point x value = 17
So y = x^2 = 17^2 = 289
So the last intersection point is (17, 289)
The focal distance, a, is given by
4ay = x^2
4(1/4)y = x^2
y = x^2
So a = 1/4
The vertex of the parabola = (0, 0)
And the focal point = (0, 0 + a) = ( 0 , 0 + 1/4) = (0, 1/4) = (0, .25)
So....now all youhave to do is to find each of the distances from the four intersection points to the poiint (0, .25)
Can you do this???? [ Note.....some technology (like WolframAlpha) might help with these calculations ]
Why hello there, Mathcounts Trainer Contest cheater, I hope you have a fine time cheating your way up the leaderboards. I wish you best luck!
Thanks! I checked it again through WolframAlpha. Its correct.