prove-that-curves-x-2-y-2-3-and-xy-2-intersect-at-the-right-angle-

Question Number 147748 by Odhiambojr last updated on 23/Jul/21

p r o v e t h a t c u r v e s x^{2} - y^{2} = 3 a n d

x y = 2 i n t e r s e c t a t t h e r i g h t a n g l e

Commented by Olaf_Thorendsen last updated on 23/Jul/21

x^{2} - y^{2} = 3 and y = \frac{2}{x}

\Rightarrow x^{2} - \frac{4}{x^{2}} = 3

x^{4} - 3 x^{2} - 4 = 0

x^{4} + x^{2} - 4 x^{2} - 4 = 0

x^{2} (x^{2} + 1) - 4 (x^{2} + 1) = 0

(x^{2} - 4) (x^{2} + 1) = 0

(x + 2) (x - 2) (x^{2} + 1) = 0

x = - 2 \Rightarrow y = \frac{2}{- 2} = - 1

x = + 2 \Rightarrow y = \frac{2}{+ 2} = + 1

\Rightarrow 2 intersection points :

A (- 2, - 1) and B (+ 2, + 1)

∙ points A and B

if y = \frac{2}{x}, \frac{d y}{d x} = - \frac{2}{x^{2}} \Rightarrow \frac{d y}{d x} ∣_{x = \pm 2} = - \frac{1}{2}

∙ point A

If x^{2} - y^{2} = 3, y = - \sqrt{x^{2} - 3}

\frac{d y}{d x} = \frac{- 2 x}{2 \sqrt{x^{2} - 3}} = - \frac{x}{\sqrt{x^{2} - 3}}

\frac{d y}{d x} ∣_{x = - 2} = - \frac{- 2}{\sqrt{4 - 3}} = + 2

∙ point B

If x^{2} - y^{2} = 3, y = + \sqrt{x^{2} - 3}

\frac{d y}{d x} = \frac{2 x}{2 \sqrt{x^{2} - 3}} = \frac{x}{\sqrt{x^{2} - 3}}

\frac{d y}{d x} ∣_{x = - 2} = \frac{+ 2}{\sqrt{4 - 3}} = + 2

The product of the slopes at A and B

is equal to - 1, (- \frac{1}{2} \times 2), that means

the two curves intercept at a right

angle .

Answered by iloveisrael last updated on 23/Jul/21

(1) x^{2} - y^{2} = 3

\Rightarrow 2 x - 2 y . y^{'} = 0

\Rightarrow m_{1} = y^{'} = \frac{x}{y}

(2) xy = 2

\Rightarrow y + {xy}^{'} = 0

\Rightarrow m_{2} = y^{'} = - \frac{y}{x}

∴ m_{1} \times m_{2} = - 1