Answer to Question #143654 in Analytic Geometry for Sarita bartwal

Question

The section of the conicoid 3x^2-y^2+z^2=1 by the plane y+2z=4 is an ellipse.
True or false with full explanation

Answer 1

This section is defined by these two equation :

$3x^2-y^2+z^2=1$

$y+2z=4$

Let's reduce the number of variables :

$y=4-2z$

$3x^2-(4-2z)^2+z^2=1$

$3x^2 -3z^2+16z-16=1$

Let's write an expression of a full square for $z$ :

$3x^2 - (3z^2-2\times \sqrt{3}z \times \frac{8}{\sqrt{3}}+\frac{64}{3})=-\frac{13}{3}$

$3(z-\frac{8}{3})^2-3x^2=\frac{13}{3}$

We clearly see that it is an equation of a hyperbola and therefore the answer to the question is NO, it is not an ellipse.