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 points

$A(-4,0),B(4,0)$

Here $\angle APB=\angle AQB=\frac{\pi }{2}$

It means $AP$ is perpendicular to $PB$

Point p lies on $y=mx+8$

So let point $P(x,mx+8)$

Slope of line $PA$ is

$m_{PA}=\frac{mx+8}{x+4}$

Slope of line $PB$ is

$m_{PB}=\frac{mx+8}{x-4}$

Line PA and PB are perpendicular SO

$m_{PA}{m}_{PB}=-1$

$\left(\frac{mx+8}{x+4}\right)\left(\frac{mx+8}{x-4}\right)=-1$

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

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

Since above equation should real and distinct root

$256m^2-4\times 48(m^2+1)>0$

$256m^2-192m^2-192>0$

$64m^2-192>0$

$m^2>3$

$m>\pm \sqrt{3}$