Let a parabola P be such that its vertex and focus lie on the positive X-axis at a distanc…

Question

Let a parabola $P$ be such that its vertex and focus lie on the positive $X$-axis at a distance $2$ and $4$ units from the origin, respectively. If tangents are drawn from $O(0,0)$ to the parabola $P$ which meet $P$ at $S$ and $R$, then the area (in sq. units) of $	riangle SOR$ is equal to

Accepted Answer

Vertex $(2,0)$, focus $(4,0)$, so $a=2$: $y^2=8(x-2)$.

Tangent: $y=m(x-2)+\dfrac2m$; through $O$: $-2m+\dfrac2m=0\Rightarrow m=\pm1$, giving lines $y=x$ and $y=-x$.

Points of contact ($a/m^2,2a/m$ in shifted coordinates, $a=2$): for $m=1$, shifted $(2,4)\to S=(4,4)$; for $m=-1$, shifted $(2,-4)\to R=(4,-4)$.

$$\text{Area}(\triangle SOR)=\frac12|x_Sy_R-x_Ry_S|=\frac12|4(-4)-4(4)|=\frac12(32)=16.$$

Solutions · 1