Answer to Question #102911 in Analytic Geometry for ah

Question #102911
Find the nature of the planar section of the conicoid x^2÷3 −y^2÷4 = z by the plane
x+2y−z = 6
1
Expert's answer
2020-02-18T05:49:07-0500

We have system of equations (conicoid and plane):{x23y24=zx+2yz=6{x23y24=zz=x+2y6{x23y24=x+2y6z=x+2y6\begin{cases} \frac{x^2}{3}-\frac{y^2}{4}=z \\ x+2y-z=6 \end{cases} \quad \begin{cases} \frac{x^2}{3}-\frac{y^2}{4}=z\\ z=x+2y-6 \end{cases} \quad \begin{cases} \frac{x^2}{3}-\frac{y^2}{4}=x+2y-6\\ z=x+2y-6 \end{cases}


x23y24=x+2y6,(x23x+34)34(y24+2y+4)+4=6\frac{x^2}{3}-\frac{y^2}{4}=x+2y-6, \quad (\frac{x^2}{3}-x+\frac{3}{4})-\frac{3}{4}-(\frac{y^2}{4}+2y+4)+4=-6


(x332)2(y2+2)2=9.25(\frac{x}{\sqrt{3}}-\frac{\sqrt{3}}{2})^2-(\frac{y}{2}+2)^2=-9.25 - equation of the hyperbolic cylinder.


{(x332)2(y2+2)2=9.25z=x+2y6\begin{cases} (\frac{x}{\sqrt{3}}-\frac{\sqrt{3}}{2})^2-(\frac{y}{2}+2)^2=-9.25 \\ z=x+2y-6 \end{cases}


Intersection of hyperbolic cylinder and plane can be:

• a line, 2 parallel lines, no points (if plane is perpendicular to z=0z=0 )

• hyperbola (if if plane isn’t perpendicular to z=0z=0 )


In our case, x+2yz=6x+2y-z=6 isn’t perpendicular to z=0z=0.

Therefore, it will be hyperbola.


Answer: hyperbola.



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