Simple Section

Geometry Level 2

The graph $x^2 + y^2 = \dfrac{z^2}{3}$ intersects with the graph $z = ky + c$ , where $k$ and $c$ are real numbers, such that the intersected points form a parabola with the point $\left (0 , \dfrac{-1}{2}, \dfrac{\sqrt{3}}{2}\right)$ as its vertex.

What is the value of $k^2$ ?

The answer is 3.

1 solution

Worranat Pakornrat
Mar 9, 2016

This is a simple demonstration of a conic section resulting in a parabola plane as shown below:

The graph $x^2 + y^2 = \dfrac{z^2}{3}$ is itself 2 conic figures (pink), and the plane $z = ky + c$ is the cutting section (blue).

Since the point $\left (0 , \dfrac{-1}{2}, \dfrac{\sqrt{3}}{2}\right)$ is the vertex of the parabola, the graph will rise along the increasing values of $z$ .

Moreover, the slope of the plane will be parallel to the cone's slant, and when taking a conic section over $x = 0$ , portraying as y-z plane, it can demonstrated as shown:

From the image, the slope of the blue plane (now a line) or $k$ is equal to the slope of the cone's slant. We know that the origin point $(0, 0, 0)$ and $\left (0 , \dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)$ (the point on the same level as the vertex) are included in this slant. Therefore, the slope = $\dfrac{\dfrac{\sqrt{3}}{2} - 0}{\dfrac{1}{2}-0}$ = $\sqrt{3}$ = $k$ .

Therefore, $k^2 = \boxed{3}$ .

Moderator note:

Good intuition for understanding the problem.

Be careful with the pictorial interpretation, since the axis are not labelled.

To find the slope, I just looked at the $yz$ -plane. The intersection of this $x=0$ plane with the cone is $y^2=\frac{z^2}3\Longleftrightarrow z=\pm\sqrt3y$ , so slope is $k=\pm\sqrt3$ and $k^2=3$ .

Laurent Shorts - 4 years, 3 months ago

Simple Section

The answer is 3.

1 solution

Moderator note:

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