Math Question and Answers


Question and Answers Forum

All Questions Topic List
Coordinate Geometry Questions
Previous in All Question Next in All Question
Previous in Coordinate Geometry Next in Coordinate Geometry
Question Number 54481 by ajfour last updated on 04/Feb/19

Commented by ajfour last updated on 04/Feb/19

$F i n d e q . o f c i r c l e a n d c o o r d i n a t e s$ $o f p o i n t s A, B, C, D i n t e r m s o f c .$
	Answered by MJS last updated on 05/Feb/19

	$A = (\begin{matrix} \sqrt{c} \\ 0 \end{matrix}) B = (\begin{matrix} p \\ p^{2} - c \end{matrix}) C = (\begin{matrix} q \\ q^{2} - c \end{matrix}) D = (\begin{matrix} \sqrt{c} \\ r \end{matrix})$ $c ⩾ 0$ $normal in B :$ $n : y = - \frac{x}{2 p} - c + p^{2} + \frac{1}{2}$ $D \in n \land ∣ D n ∣= r = - \frac{\sqrt{c}}{2 p} - c + p^{2} + \frac{1}{2}$ $\Rightarrow p^{5} - 2 {c p}^{3} + \sqrt{c} p^{2} + c (c - 2) p + \sqrt{c^{3}} = 0$ ${(p - \sqrt{c})}^{2} (p^{3} + 2 \sqrt{c} p^{2} + c p + \sqrt{c}) = 0$ $p^{3} + 2 \sqrt{c} p^{2} + c p + \sqrt{c} = 0$ $p = z - \frac{2 \sqrt{c}}{3}$ $z^{3} - \frac{c}{3} z - \frac{(2 c - 27) \sqrt{c}}{27} = 0$ $this has got 1 real solution for 0 ⩽ c < \frac{27}{4},$ $2 real solutions for c = \frac{27}{4}, 3 real solutions for c > \frac{27}{4}$ $\Rightarrow we can calculate p \Rightarrow r, q for given c$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum