Geometricus Geometriorum (HP I, philosophical stone)

Geometry Level 4

( x 2017 ) 2 576 + ( y 2016 ) 2 625 = 1 \frac{(x - 2017)^2}{576} + \frac{(y - 2016)^2}{625} = 1

Given the ellipse above, let

  • the distance between the foci of this ellipse be a a ;
  • the sum of the distances from the point ( 2017 + 24 cos π 4 , 2016 + 25 sin π 4 ) \left(2017 + 24\cos \dfrac{\pi}{4}, \ 2016 + 25\sin \dfrac{\pi}{4} \right) to the foci be b b .

Find a + b a + b .


The answer is 64.

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1 solution

The ellipse equation shows that the minor radius and major radius are α = 576 = 24 \alpha = \sqrt{576} = 24 and β = 625 = 25 \beta = \sqrt{625} = 25 respectively.

The distance between a focus and the centre of the ellipse is given by f = β 2 α 2 = 625 576 = 49 = 7 f = \sqrt{\beta^2 - \alpha^2} = \sqrt{625-576} = \sqrt{49} = 7 , therefore, the distance between the two foci a = 2 × 7 = 14 a = 2 \times 7 = 14 . ( See under Focus )

Let x x and y y be the coordinates of the given point P P , then we have:

L H S = ( 2017 + 24 cos π 4 2017 ) 2 576 + ( 2016 + 25 sin π 4 2016 ) 2 625 = 1 2 + 1 2 = 1 = R H S \begin{aligned} LHS & = \frac {\left(2017 + 24 \cos \frac \pi 4 - 2017\right)^2}{576} + \frac {\left(2016 + 25 \sin \frac \pi 4 - 2016\right)^2}{625} = \frac 12 + \frac 12 = 1 = RHS \end{aligned}

This means that point P P is on the ellipse, and the sum of distances of any point on the ellipse to the foci is twice the major radius, that is b = 2 β = 2 × 25 = 50 b = 2\beta = 2 \times 25 = 50 . ( See under Focus )

a + b = 14 + 50 = 64 \implies a+b = 14+50 = \boxed{64} .

Exactly, (+1)... Thank you, Chew.

Guillermo Templado - 4 years, 9 months ago

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