If the area of the triangle whose one vertex is at the vertex of the parabola y^2+4(x-a^2)…

Question

If the area of the triangle whose one vertex is at the vertex of the parabola $y^2+4(x-a^2)=0$ and the other two vertices are the points of intersection of the parabola and the $y$-axis, is $250$ sq. units, then a value of $a$ is:

Accepted Answer

At $x=0$: $y^2=4a^2\Rightarrow y=\pm2a$. Vertex $(a^2,0)$; base $=4a$, height $=a^2$:

$$\text{Area}=\frac12(4a)(a^2)=2a^3=250\Rightarrow a^3=125\Rightarrow a=5.$$

Solutions · 1