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# Parabolas Practice Problem - MATHCOUNTS

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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.

Oct 14, 2019

#1
+109563
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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 ]

Oct 14, 2019
edited by CPhill  Oct 14, 2019
#2
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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!

Oct 15, 2019
#3
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Thanks! I checked it again through WolframAlpha. Its correct.

Oct 15, 2019
#4
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I had this on Alcumus...I remember doing it!

Oct 16, 2019