6
Question
Show that the equation to the circle circumscribing the triangle formed by the three tangents to the parabola r=2a1−cosθ drawn at the points whose vectorial angles are α,β and γ, is
r=acosecα2cosecβ2cosecγ2sin((α+β+γ)2−θ),
and hence that it always passes through the focus.
r=acosecα2cosecβ2cosecγ2sin((α+β+γ)2−θ),
and hence that it always passes through the focus.
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Solution
Equation to parabola is given by
r=2a1−cosθ or 2ar=1−cosθ
Let P, Q, R be the points on the parabola where tangents are drawn. The vectorial angle of P, Q, R are given to be α,β and γ respectively. Then the equation to tangents at P and Q are respectively.
2ar=cos(θ−α)−cosθ.....1
and 2ar=cos(θ−β)−cosθ.....2
and cos(θ−α)=cos(θ−β)
To get the point of intersection of 1 and 2, we solve them. So subtracting 2 from 1, we get
or θ−α=−(θ−β)
as α≠β
2θ=α+β
θ=α+β2
Putting this value in 1 we get
r=2acosβ−α2−cosα+β2
Hence if ABC be the vertices of the circumscribing triangle, the co ordinates of A will be
⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩2acosβ−α2−cosα+β2,α+β2⎫⎪
⎪
⎪⎬⎪
⎪
⎪⎭={acscα2cscβ2,α+β2}
Similarly, the other vertices B and C will be
{acscβ2,cscγ2,cscβ+γ2} and {acscγ2,cscα2,cscα+γ2}
We see that the co ordinates of A, B and C be given by
(ksinλ2cscΦ−λ2) where k=acscα2,cscβ2,cscγ2
and Φ=α+β+γ2
Hence the vertices of r=ksin(Φ−θ) lie on the curve
or △ABC is inscribed in the curve
r=acscα2cscβ2cscγ2sin(α2+β2+γ2−θ)
This is clearly a circle passing through the focus
r=2a1−cosθ or 2ar=1−cosθ
Let P, Q, R be the points on the parabola where tangents are drawn.
The vectorial angle of P, Q, R are given to be α,β and γ respectively.
Then the equation to tangents at P and Q are respectively.
2ar=cos(θ−α)−cosθ.....1
and 2ar=cos(θ−β)−cosθ.....2
and cos(θ−α)=cos(θ−β)
To get the point of intersection of 1 and 2, we solve them.
2ar=cos(θ−α)−cosθ.....1
and 2ar=cos(θ−β)−cosθ.....2
and cos(θ−α)=cos(θ−β)
To get the point of intersection of 1 and 2, we solve them.
So subtracting 2 from 1, we get
or θ−α=−(θ−β)
as α≠β
2θ=α+β
θ=α+β2
Putting this value in 1 we get
r=2acosβ−α2−cosα+β2
Hence if ABC be the vertices of the circumscribing triangle, the co ordinates of A will be
⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩2acosβ−α2−cosα+β2,α+β2⎫⎪ ⎪ ⎪⎬⎪ ⎪ ⎪⎭={acscα2cscβ2,α+β2}
Similarly, the other vertices B and C will be
{acscβ2,cscγ2,cscβ+γ2} and {acscγ2,cscα2,cscα+γ2}
We see that the co ordinates of A, B and C be given by
(ksinλ2cscΦ−λ2) where k=acscα2,cscβ2,cscγ2
and Φ=α+β+γ2
Hence the vertices of r=ksin(Φ−θ) lie on the curve
or △ABC is inscribed in the curve
r=acscα2cscβ2cscγ2sin(α2+β2+γ2−θ)
This is clearly a circle passing through the focus
or θ−α=−(θ−β)
as α≠β
2θ=α+β
θ=α+β2
Putting this value in 1 we get
r=2acosβ−α2−cosα+β2
Hence if ABC be the vertices of the circumscribing triangle, the co ordinates of A will be
⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩2acosβ−α2−cosα+β2,α+β2⎫⎪ ⎪ ⎪⎬⎪ ⎪ ⎪⎭={acscα2cscβ2,α+β2}
Similarly, the other vertices B and C will be
{acscβ2,cscγ2,cscβ+γ2} and {acscγ2,cscα2,cscα+γ2}
We see that the co ordinates of A, B and C be given by
(ksinλ2cscΦ−λ2) where k=acscα2,cscβ2,cscγ2
and Φ=α+β+γ2
Hence the vertices of r=ksin(Φ−θ) lie on the curve
or △ABC is inscribed in the curve
r=acscα2cscβ2cscγ2sin(α2+β2+γ2−θ)
This is clearly a circle passing through the focus
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