bisector theorem , cosine rule or other method?

Question

A fixed point $A(a,0)$, where $a>0$ and a straight line $l$ is given $x = -1$, $B$ is a moving point on the second quadrant and lies on the straight line $l$, the angle bisector of $\angle BOA$ intersects $AB$ at point $C$.

(a) Find the locus of point $C$ and state the range values of $x$ and $y$.

(b) Discuss the relationship between the values of $a$ and the type of the curve of the equation obtained in (a).

I have tried to use cosine rule to and bisector theorem to obtained the locus. but the equation seem too complicated. Any other good method please share with me !!

Answer 1

Let $B=(-1,t)$, $C=(x,y)$.

Then $$\frac{y}{x-a}=\frac{t}{-1-a}$$

Angle AOB can be obtained to be $$\cos AOB=\frac{-a}{a\sqrt{1+t^2}}=\frac{-1}{\sqrt{1+t^2}}$$

Using $$\cos^2 AOC = \frac{1+\cos AOB}{2}$$ we have $$\frac{y^2}{x^2}=\tan^2 AOC = \sec^2 AOC - 1=\frac{2}{1-\frac{1}{\sqrt{1+t^2}}}-1=\frac{2\sqrt{1+t^2}}{\sqrt{1+t^2}-1}-1=\frac{\sqrt{1+t^2}+1}{\sqrt{1+t^2}-1}=\frac{\left(\sqrt{1+t^2}+1\right)^2}{t^2}$$

$$\frac{y}{x}=\frac{\sqrt{1+t^2}+1}{t}$$

$$\left(t\frac{y}{x}-1\right)^2=1+t^2$$ $$\left(ty-x\right)^2=\left(1+t^2\right)x^2$$ $$\left(\frac{(1+a)y^2}{x-a}+x\right)^2=\left(1+\frac{(1+a)^2y^2}{(x-a)^2}\right)x^2$$ $$((1+a)y^2+x(x-a))^2=((x-a)^2+(1+a)^2y^2)x^2$$ $$(1+a)^2y^4+2(1+a)x(x-a)y^2=(1+a)^2x^2y^2$$ $$y^2+\frac{2}{1+a}x(x-a)-x^2=0$$ $$\left(\frac{2}{1+a}-1\right)x^2-\frac{2a}{1+a}x+y^2=0$$ $$(1-a)x^2-2ax+(1+a)y^2=0$$

If $0<a<1$, then it is an ellipse.

If $a=1$, then it is a parabola.

If $a>1$, then it is a hyperbola.

Answer 2

Point $C$ must have the same distance from $x$ axis and from line $OB$

It must also lie on the segment $AB$.

$A(a,\;0)$ and point $B$ lies on the line $x=-1$ so its coordinates are $B(-1,\;s)$ for any $s$

Line $AB$ has equation $y=-\dfrac{s}{a+1}(x-a)$ and line $AC$ has equation $y=-s x$

$sx +y =0$

$C$ coordinates are $C\left(t,\;-\dfrac{s}{a+1}(t-a)\right)$, parametric form of the equation of the line $AB$ must be equal to the distance from $C$ to the $x$ axis, that is the $y$ coordinate of $C$

$$\dfrac{\left|s t -\dfrac{s}{a+1}(t-a)\right|}{\sqrt{s^2+1}}=\left|-\dfrac{s}{a+1}(t-a)\right|$$

$y$ coordinate of $C$ is $y=-\dfrac{s}{a+1}(t-a)\to s=\dfrac{(a+1) y}{a-t}$

squaring we get $$\frac{\left(s t-\frac{s (t-a)}{a+1}\right)^2}{s^2+1}=\frac{s^2 (t-a)^2}{(a+1)^2}$$ Plug the value of $s$ and remember that $t=x$, simplify and finally $$(1-a) x^2-2 a x+(a+1) y^2=0$$ is the equation of the locus which can be an ellipse if $0<a<1$, an hyperbola if $a> 1$ or a parabola if $a=1$

Hope this helps

Answer 3

Let $B(-1,t)$ and $\alpha =$ angle $AOC$. We have $\tan \alpha =y/x$ and $\tan 2\alpha = -t $ . Use the double angle formula to calculate

$$-t=\frac{2 \tan \alpha }{1-\tan ^2\alpha }=\frac{2y/x}{1-y^2/x^2}$$

On the other hand, point $C$ on $AB$ gives

$$\frac{y}{t}=\frac{a-x}{a+1}$$

Eliminating $t$ we get the equation of the locus

$$(a+1)\left(y^2-x^2\right)=2x(a-x)$$