Question

Find the nature of the planar section of the conicoid x^2/3 −y^2/4 = z by the plane x+2y−z = 6

Answer 1

$\dfrac{x^2}{3}-\dfrac{y^2}{4}=z$

$x+2y−z=6$

We will find z

$z=x+2y−6$

$4 x^ 2 ​ − 3 y^ 2 ​ =12(x+2y−6)$

$4(x^ 2 −3x)−3(y^ 2 +8y)=−72$

$4(x−\frac{3}{2}) ^2 −3(y+4) ^2 =−72+9−48=−111$

$4 \frac{ ​ (x−\frac 32 ​ )^2}{111} ​ − 3 \frac{ ​ (y+4)^ 2}{111} ​ =−1$

$\frac{ ​ (x−\frac 32 ​ )^2}{\frac{111}{4}} ​ − \frac{ ​ (y+4)^ 2}{\frac{111}{3}} ​ =−1$

This is a equation of conjugated hyperbola

$center( \frac 32 ​ ,−4)$

semi−axes=

$a= \sqrt{ 111/4 ​} ​ ,b= \sqrt{ 111/3 ​ } ​$