Question-17963

Question Number 17963 by ajfour last updated on 13/Jul/17

Commented by ajfour last updated on 13/Jul/17

AB = 2 a, CD = 2 b

Find r in terms of a, b, ϕ, and R .

Hence find {AP}^{2} + {BP}^{2} + {CP}^{2} + {DP}^{2} .

Commented by ajfour last updated on 13/Jul/17

Commented by ajfour last updated on 13/Jul/17

x = \sqrt{R^{2} - a^{2}}, y = \sqrt{R^{2} - b^{2}}

\sin ϕ_{1} = \frac{x}{r}, \sin ϕ_{2} = \frac{y}{r}

ϕ_{1} + ϕ_{2} = ϕ, so

\sin^{- 1} (\frac{\sqrt{R^{2} - a^{2}}}{r}) + \sin^{- 1} (\frac{\sqrt{R^{2} - b^{2}}}{r}) = ϕ \dots (i)

cannot r be obtained explicitly ?

AP = a - PM, BP = a + PM

CP = b - NP, DP = b + NP

{AP}^{2} + {BP}^{2} + {CP}^{2} + {DP}^{2} =

{2 a}^{2} + {2 PM}^{2} + {2 b}^{2} + {2 NP}^{2} \dots (ii)

Now, {PM}^{2} + {NP}^{2} = r^{2} - x^{2} + r^{2} - y^{2}

and with x^{2} = R^{2} - a^{2}, y^{2} = R^{2} - b^{2}

we have

{PM}^{2} + {NP}^{2} = {2 r}^{2} - (x^{2} + y^{2})

= {2 r}^{2} - ({2 R}^{2} - a^{2} - b^{2})

= {2 r}^{2} + a^{2} + b^{2} - {2 R}^{2}

substituting in (ii), we get

{AP}^{2} + {BP}^{2} + {CP}^{2} + {DP}^{2}

= {(2 a)}^{2} + {(2 b)}^{2} - 4 (R^{2} - r^{2});

r is obtained from (i) .

Commented by b.e.h.i.8.3.417@gmail.com last updated on 14/Jul/17

n i c e a n d b e a u t i f u l m r A j f o u r .