Let-be-the-hyperbola-xy-b-b-0-and-P-be-a-point-on-Let-Q-be-the-image-of-reflection-of-P-about-the-origin-Construct-a-circle-centred-at-P-with-radius-PQ-cuts-at-the-points-B-C-

Question Number 223487 by Tawa11 last updated on 26/Jul/25

Let Φ be the hyperbola xy = b², b ≠ 0, and
P be a point on Φ.
Let Q be the image of reflection of P about
the origin. Construct a circle ω centred at P
with radius PQ.

ω cuts Φ at the points B, C, D, Q.
Prove that ΔBCD is equilateral, no matter
what the value of b is.

Answered by mr W last updated on 27/Jul/25

Commented by mr W last updated on 28/Jul/25

P(p,(b^2 /p)) R=PQ=2×OP=2(√(p^2 +(b^4 /p^2 )))=2p(√(1+(b^4 /p^4 ))) x=p+R cos θ, y=(b^2 /p)+R sin θ xy=b^2 (p+R cos θ)((b^2 /p)+R sin θ)=b^2 (1+2(√(1+(b^4 /p^4 ))) cos θ)((b^2 /p^2 )+2(√(1+(b^4 /p^2 ))) sin θ)=(b^2 /p^2 ) with λ=(b^2 /p^2 )=tan α (1+2(√(1+λ^2 )) cos θ)(λ+2(√(1+λ^2 )) sin θ)=λ (λ/( (√(1+λ^2 )))) cos θ+(1/( (√(1+λ^2 )))) sin θ+2 sin θ cos θ=0 sin α cos θ+cos α sin θ+sin 2θ=0 sin (θ+α)=−sin 2θ case 1: 2θ=θ+α+(2k+1)π ⇒θ_4 =α+(2k+1)π ⇒point Q case 2: θ+α+2θ=2kπ ⇒θ_(1,2,3) =−(α/3)+((2kπ)/3) ⇒point B,C,D PB, PC, PD are at angle 120° to each other, i.e. ΔBCD is equilateral. this is independent from λ, i.e. for any value of b and any position of P.

$${P}\left({p},\frac{{b}^{\mathrm{2}} }{{p}}\right) \\ $$$${R}={PQ}=\mathrm{2}×{OP}=\mathrm{2}\sqrt{{p}^{\mathrm{2}} +\frac{{b}^{\mathrm{4}} }{{p}^{\mathrm{2}} }}=\mathrm{2}{p}\sqrt{\mathrm{1}+\frac{{b}^{\mathrm{4}} }{{p}^{\mathrm{4}} }} \\ $$$${x}={p}+{R}\:\mathrm{cos}\:\theta,\:{y}=\frac{{b}^{\mathrm{2}} }{{p}}+{R}\:\mathrm{sin}\:\theta \\ $$$${xy}={b}^{\mathrm{2}} \\ $$$$\left({p}+{R}\:\mathrm{cos}\:\theta\right)\left(\frac{{b}^{\mathrm{2}} }{{p}}+{R}\:\mathrm{sin}\:\theta\right)={b}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\frac{{b}^{\mathrm{4}} }{{p}^{\mathrm{4}} }}\:\mathrm{cos}\:\theta\right)\left(\frac{{b}^{\mathrm{2}} }{{p}^{\mathrm{2}} }+\mathrm{2}\sqrt{\mathrm{1}+\frac{{b}^{\mathrm{4}} }{{p}^{\mathrm{2}} }}\:\mathrm{sin}\:\theta\right)=\frac{{b}^{\mathrm{2}} }{{p}^{\mathrm{2}} } \\ $$$${with}\:\lambda=\frac{{b}^{\mathrm{2}} }{{p}^{\mathrm{2}} }=\mathrm{tan}\:\alpha \\ $$$$\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{cos}\:\theta\right)\left(\lambda+\mathrm{2}\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{sin}\:\theta\right)=\lambda \\ $$$$\frac{\lambda}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}\:\mathrm{cos}\:\theta+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}\:\mathrm{sin}\:\theta+\mathrm{2}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\mathrm{sin}\:\alpha\:\mathrm{cos}\:\theta+\mathrm{cos}\:\alpha\:\mathrm{sin}\:\theta+\mathrm{sin}\:\mathrm{2}\theta=\mathrm{0} \\ $$$$\mathrm{sin}\:\left(\theta+\alpha\right)=−\mathrm{sin}\:\mathrm{2}\theta \\ $$$${case}\:\mathrm{1}:\:\mathrm{2}\theta=\theta+\alpha+\left(\mathrm{2}{k}+\mathrm{1}\right)\pi \\ $$$$\Rightarrow\theta_{\mathrm{4}} =\alpha+\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\:\:\:\Rightarrow{point}\:{Q} \\ $$$${case}\:\mathrm{2}:\:\theta+\alpha+\mathrm{2}\theta=\mathrm{2}{k}\pi \\ $$$$\Rightarrow\theta_{\mathrm{1},\mathrm{2},\mathrm{3}} =−\frac{\alpha}{\mathrm{3}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\:\Rightarrow{point}\:{B},{C},{D} \\ $$$${PB},\:{PC},\:{PD}\:{are}\:{at}\:{angle}\:\mathrm{120}° \\ $$$${to}\:{each}\:{other},\:{i}.{e}.\:\Delta{BCD}\:{is}\: \\ $$$${equilateral}. \\ $$$${this}\:{is}\:{independent}\:{from}\:\lambda,\:{i}.{e}. \\ $$$${for}\:{any}\:{value}\:{of}\:{b}\:{and}\:{any}\:{position}\: \\ $$$${of}\:{P}. \\ $$

Commented by mr W last updated on 27/Jul/25