Question

if A(a,0),B(-a,0) and angle APB =45° then locus of P is

Answer 1

$\textbf{Given:}$

$\text{A(a,0), B(-a,0) and \angle{APB}=45^{\circ} }$

$\textbf{To find:}$

$\text{Locus of P}$

$\textbf{Solution:}$

$\text{Let the moving point be P(h,k) }$

$\text{Slope of AP}=\dfrac{y_2-y_1}{x_2-x_1}$

$\text{Slope of AP}=\dfrac{k-0}{h-a}$

$\text{Slope of AP,}\,m_1=\dfrac{k}{h-a}$

$\text{Slope of BP,}\,m_2=\dfrac{k}{h+a}$

$\text{Since angle between AP and BP is 45^{\circ} ,}$

$\implies|\dfrac{m_1-m_2}{1+m_1\,m_2}|=tan\,45^{\circ}$

$\implies|\dfrac{\frac{k}{h-a}-\frac{k}{h+a}}{1+(\frac{k}{h-a})(\frac{k}{h+a})}|=1$

$\implies|\dfrac{\frac{k(h+a)-k(h-a)}{(h-a)(h+a)}}{1+\frac{k^2}{h^2-a^2}}|=1$

$\implies|\dfrac{\frac{kh+ak-kh+ak}{h^2-a^2}}{1+\frac{k^2}{h^2-a^2}}|=1$

$\implies|\dfrac{\frac{2ak}{h^2-a^2}}{\frac{h^2-a^2+k^2}{h^2-a^2}}|=1$

$\implies|\dfrac{2ak}{h^2-a^2+k^2}|=1$

$\implies\dfrac{2ak}{h^2-a^2+k^2}=\pm\,1$

$\text{case(1):}$

$\implies\dfrac{2ak}{h^2-a^2+k^2}=1$

$\implies\,h^2-a^2+k^2=2ak$

$\implies\,h^2+k^2-2ak-a^2=0$

$\text{case(2):}$

$\implies\dfrac{2ak}{h^2-a^2+k^2}=-1$

$\implies\,h^2-a^2+k^2=-2ak$

$\implies\,h^2+k^2+2ak-a^2=0$

$\textbf{Answer:}$

$\textbf{The locus of P are \bf\,x^2+y^2-2ay-a^2=0 and \bf\,x^2+y^2+2ay-a^2=0 }$

$\textbf{Note:}$

$\text{sometimes the problem can be done by without using modulus}$