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Let P be a variable point on the parabola y2=2x and Let B and C be points on the y-axis so that the circle (x−1)2+y2=1 is inscribed in triangle PBC. Find the minimum area of triangle PBC.
 

 Nov 22, 2020

4+7 Answers

 #2
avatar+118702 
+1

 

Graphically I have found the answer to be about 8

 

Here is my interactive graph.  You can move the point P around to get different areas.

 

https://www.geogebra.org/classic/mw7schnm

 

I have not worked out how to do it algebraically yet.

 Nov 30, 2020
 #6
avatar+130477 
+1

I saw this earlier......but I have no clue.....

 

I don't know if your answer is correct, or not, Melody.....but it certainly seems to  be  close

 

Nice work in Geogebra, anyway  smileysmileysmiley

 

 

cool cool cool

 Dec 1, 2020
 #8
avatar+118702 
+2

Thanks Chris, I have fun sketching these. 

 

My sketch is accurate, my answer would be right within approx 2 decimal places.

My answer has as much accuracy as the Geogebra program allows. This appears to be 2 dp.

Melody  Dec 1, 2020
edited by Melody  Dec 1, 2020
 #7
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I believe melodies answer is correct, as I have experimented and no other triangle can surpass the minisculity of the triangle that beholds an area of 8. Algebraicly I was quite confuzzled but the graph is undoubtedly correct.

 Dec 1, 2020

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