Question

Set of values of $m$ for which two points P & Q lies on the line $y=mx+8$ so that $\angle APB=\angle AQB=\frac{\pi }{2}$, where $A\equiv (-4,0),B\equiv (4,0)$ is

• A. $(-\infty ,-\sqrt{3})\cup (\sqrt{3},\infty )$
• B. $[-\sqrt{3},\sqrt{3}]$
• C. $(-\infty ,-1)\cup (1,\infty )$
• D. $\{-\sqrt{3},\sqrt{3}\}$

Answer 1

The correct option is A $(-\infty ,-\sqrt{3})\cup (\sqrt{3},\infty )$

Given : $y=mx+8\angle APB=\angle AQB=\frac{\pi }{2}$

if $\angle APB=90°$

$\therefore \text{slope of}AP\times \text{slope of}B=-1$

$\text{Let}P(x,y)$

Slope of $AP=\frac{y-0}{x-(-4)}=\frac{y}{x+4}$

$BP=\frac{y-0}{x-4}=\frac{y}{x-4}$

$\text{slope}\left(AP\right)\times \text{slope}\left(BP\right)=-1$

$\frac{y}{x+4}\times \frac{y}{x-4}=-1$

$y^2=-({x-16)}^2⟶(1)$

$\because y=mx+8$

$y^2={(mx+8)}^2⟶\left(2\right)$

from $1$ & $2$,

$-x^2+16=m^2{n}^2+64+2mx8$

$x^2(m^2+1)+2mx8+48=0$

$x^2(m^2+1)+16mx8+48=0$

$D\ge 0$

${16m}^2-4(m^2+1)48>0$

$64m^2-192>0$

$m^2-3>0$

$m\in (-\infty ,-\sqrt{3})U(\sqrt{3},\infty )$