The tangents at the points A(1,3) and B(1,-1) on the parabola y^2-2x-2y=1 meet at the poin…

Question

The tangents at the points $A(1,3)$ and $B(1,-1)$ on the parabola $y^2-2x-2y=1$ meet at the point $P$. Then the area (in unit$^2$) of the triangle $PAB$ is:

Accepted Answer

$(y-1)^2=2(x+1)$; with $X=x+1,Y=y-1$ this is $Y^2=2X$. $A\to(2,2)$, $B\to(2,-2)$.

Tangent at $(X_1,Y_1)$: $YY_1=X+X_1$. At $A$: $2Y=X+2$; at $B$: $-2Y=X+2$. Intersection: $Y=0,X=-2\Rightarrow P=(-3,1)$.

$AB$ is vertical, length $4$ (from $y=-1$ to $y=3$); horizontal distance from $P$ to line $x=1$ is $4$.

$$\text{Area}=\frac12\cdot4\cdot4=8.$$

Solutions · 1