If the line $x+y-1=0$ is a tangent to a parabola with focus (1,2) at A and intersects the directrix at B and tangent...

Question

If the line $x+y-1=0$ is a tangent to a parabola with focus (1,2) at A and intersects the directrix at B and tangent at vertex C respectively, then find the value of AC.BC

First of, they have given absolutely no infomation regarding the parabola, leaving me with no choice but to assume some standard case

If it is of the form $y^2=4ax$, the axis of the parabola is $y=2$

So $$(y-2)^2=4a(x-x_1)$$

The equation of tangent to the parabola is $$y-2=m(x-x_1)+\frac am$$ Putting m=-1

$$y-2=x_1-x-a$$ $$x+y=2+x_1-a$$

So $$2+x_1-a=1$$ $$x_1=a-1$$

The distance from vertex to focus is +a So
As given $$x_1+a=1$$ x coordinate of focus given $$(a-1)+a=1$$ $$a=1$$ Then $$x_1=0$$ Now vertex=(0,2) U can try now finding directrix and its intersection point with tangent and intersection point with tangent at vertex then u get

                AC= square root of 2 
                BC=square root of 2 
   Now,
        AC.BC=2

That’s all I could do.

Answer 1

A partial answer.

The parabola with focus $F=(1,2)$ and directrix $lx+my+l=0$ has equation $$(l^2+m^2)((x-1)^2+(y-2)^2)-(lx+my+l)^2=0$$ and experimenting in geogebra I found it happens to meet $y+x-1=0$ doubly in $A$: $(x,y)=(\frac{l-m}{l+m},\frac{2m}{l+m})$ (substitute the line $y=1-x$ in the equation for the parabola and check that it factors as $(mx+lx+m-l)^2$) so it's tangent there for all $l,m$ with $l+m\neq 0$.

The directrix $lx+my+l=0$ meets $y+x-1=0$ in $B:(x,y)=(-\frac{l+m}{l-m},\frac{2l}{l-m})$.

The axis is parallel to $ -mx +ly$ and passes through the focus, and intersects the parabola in the vertex which is $V:(x,y)=(\frac{m(m-l)}{l^2+m^2},\frac{m^2-lm+2l^2}{l^2+m^2})$ so the tangent at $V$ (which is parallel to $lx+my=0$ and goes through $V$) is $lx+my-m=0$ which meets $y+x-1=0$ in $C: (x,y)=(0,1)$.

Now $$AC\cdot BC=\sqrt{\left(\frac{l-m}{l+m}-0\right)^2+\left(\frac{2\,m}{l+m}-1\right)^2}\sqrt{\left(-\frac{l+m}{l-m}-0\right)^2+\left(\frac{2\,l}{ l-m}-1\right)^2}=\sqrt{2\frac{(l-m)^2}{(l+m)^2}}\sqrt{2\frac{(l+m)^2}{(l-m)^2}}=2$$ for $l\pm m\neq 0.$ When $l=m$ the points $A$ and $C$ coincide and $B$ goes off to infinity. When $l=-m$ the parabola degenerates to a double line $(x+y-3)^2=0$; which is not tangent to $x+y-1=0$ in the finite plane; $A$ goes off to infinity with $B$ and $C$ coinciding.