Question #143654
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
1
Expert's answer
2020-11-16T19:48:49-0500

This section is defined by these two equation :

3x2y2+z2=13x^2-y^2+z^2=1

y+2z=4y+2z=4

Let's reduce the number of variables :

y=42zy=4-2z

3x2(42z)2+z2=13x^2-(4-2z)^2+z^2=1

3x23z2+16z16=13x^2 -3z^2+16z-16=1

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

3x2(3z22×3z×83+643)=1333x^2 - (3z^2-2\times \sqrt{3}z \times \frac{8}{\sqrt{3}}+\frac{64}{3})=-\frac{13}{3}

3(z83)23x2=1333(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.


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