# Find $k$ such that the intersection of $x+ky=1$ and $y^2 - x^2 - z^2 = 1$ is an ellipse or a hyperbola

Find the values of $k$ such that the intersection of the plane $x+ky=1$ with the two-sheeted elliptic hyperboloid $y^2 - x^2 - z^2 = 1$ is (a) an ellipse and (b) a hyperbola.

My attempt is the following:

If I choose a particular value for $x$ such that $x = 1 - ky$, and substitute $x = 1 - ky$ into the formula for the hyperboloid, I get:

$$y^2 - (1-ky)^2 - z^2 = 1$$

which is equivalent to:

$$(k^2-1)y^2 - 2ky + z^2 + 2 = 0$$.

This equation either represents a conic section or a degenerate conic section. For it to represent an ellipse, we need that $k^2-1>0$, meaning that $|k|>1$. Similarly, to obtain the values of $k$ such that the equation represents a hyperbola, we need $k^2-1<0$, meaning that $|k|<1$. This matches the answers in the back of the book I'm using.

But there is a problem with this solution; I'm not sure if it is correct.

The problem is that the equation that I found above is in terms of $y$ and $z$; but what exactly does it represent? It seems to correspond to a plane curve on the yz-plane only, but the intersection that I'm looking for is not necessarily confined to the yz-plane. Analogously, f I substituted $y=(1-x)/k$ instead of substituting $x = 1-ky$ into the equation for the hyperboloid, I would obtain an equation in terms of x and z only. So, what exactly does this equation that I found represent?

## Update

I think that the expression $(k^2-1)y^2 - 2ky + z^2 + 2 = 0$ essentially represents the projection of the intersection onto the $yz$ plane, and it is an ellipse for $|k|>1$. But how do I go from here to showing that the whole intersection is an ellipse when $|k|>1$, or a hyperbola when $|k|<1$?

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Note: Your intersection is not confined to the yz plane, even though you finally obtained an equation (through the process of substation and elimination) in only $y$ and $z$. You still have to substitute it back into $x+ky=1$ to obtain the $x$ coordinate, and this will likely not be 0. Compare this to solving a 3-variable linear equation, where you eliminate for x and then y, leaving $z=a$ for some value, you will still need to substitute back to find the values of $y$ and $x$. –  Calvin Lin Jan 6 '13 at 0:14