Question

Let $A(ae,0),B(-ae,0)$ are two points. The equation to the locus of $P$ such that $PA-PB=2a$, is

• A. $\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1$
• B. $\frac{x^2}{a^2}-\frac{y^2}{a^2(1-e^2)}=1$
• C. $\frac{x^2}{a^2}+\frac{y^2}{a^2(1+e^2)}=1$
• D. $\frac{x^2}{a^2}-\frac{y^2}{a^2(1+e^2)}=1$

Answer 1

The correct option is A $\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1$

$(PA{)}^2+(PB{)}^2-2\left(PA\right)\left(PB\right)=4a^2$
$(n-ae{)}^2+k^2+(h+ae{)}^2+k^2=4a^2+2\left(PA\right)\left(PB\right)$
$(h^2+a^2{e}^2+k^2-2aeh)+k^2+h^2+a^2{e}^2+2hae=4a^2+\sqrt{h^2+a^2{e}^2+k^2-2aeh\times (h^2+k^2+a^2{e}^2+2aeh)}$
$\frac{h^2}{a^2}+\frac{k^2}{a^2(1-e^2)}=1$
$\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1$