Foci of sliding of ellipse

Geometry Level 3

An ellipse with major axis 4 and minor axis 2 touches both the coordinate axes and slides between them(tangent to both) then locus of its focus is

(A hint in title it's plural)

(x²+y²)(1-x²y²)=16x²y² (x²-y²)(1+x²y²)=16x²y² (x²-y²)(1-x²y²)=16x²y² (x²+y²)(1+x²y²)=16x²y²

Akshat Kumar by Akshat Kumar , 19, India

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1 solution

Michael Mendrin
Aug 7, 2018

We can use the parametric equations for an ellipse of semi-major and semi-minor axes a , b a, b , rotated by θ \theta , with one focus at the origin ( 0 , 0 ) (0,0)

x ( t ) = b 2 C o s ( t − θ ) a + a 2 − b 2 C o s ( t ) x(t)=\dfrac { b^2 Cos(t-\theta) } { a+\sqrt{a^2-b^2}Cos(t) }

y ( t ) = b 2 S i n ( t − θ ) a + a 2 − b 2 C o s ( t ) y(t)=\dfrac { b^2 Sin(t-\theta) } { a+\sqrt{a^2-b^2}Cos(t) }

To find the distance of the horizontal and vertical tangents from the origin, we solve the following equations for t t

x ′ ( t ) = 0 x'(t)=0
y ′ ( t ) = 0 y'(t)=0

so that we end up with the distances as a function of θ \theta

X ( θ ) = b 2 a 2 + ( a 2 − b 2 ) S i n ( θ ) + ( a 2 − b 2 ) C o s ( θ ) X(\theta)=\dfrac{b^2}{\sqrt{ a^2+(a^2-b^2)Sin(\theta) } + (\sqrt{ a^2-b^2 }) Cos(\theta) }

Y ( θ ) = b 2 a 2 + ( a 2 − b 2 ) C o s ( θ ) + ( a 2 − b 2 ) S i n ( θ ) Y(\theta)=\dfrac{b^2}{\sqrt{ a^2+(a^2-b^2)Cos(\theta ) } + (\sqrt{ a^2-b^2 }) Sin(\theta) }

which satisfies the equation

( X 2 + Y 2 ) ( b 4 + X 2 Y 2 ) = 4 a 2 X 2 Y 2 (X^2+Y^2)(b^4+X^2Y^2)=4a^2X^2Y^2

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