Answer to Question #104624 in Analytic Geometry for Deepak

Question #104624
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
2020-03-05T16:52:35-0500

x23y24=z\dfrac{x^2}{3}-\dfrac{y^2}{4}=z

x+2yz=6x+2y−z=6

We will find z

z=x+2y6z=x+2y−6


x23y24=x+2y6\dfrac{x^2}{3}-\dfrac{y^2}{4}=x+2y−6

4x23y212x24y=724(x23x)3(y2+8y)=724x^2-3y^2-12x-24y=-72\\ 4(x^2-3x)-3(y^2+8y)=-72\\

4(x3)23(y+2)2=72+948=1114(x-3)^2-3(y+2)^2=-72+9-48=-111\\


4(x32)1113(y+4)2111=1(x32)1114(y+4)21113=1\dfrac{4(x-\dfrac{3}{2})}{111}-\dfrac{3(y+4)^2}{111}=-1\\ \dfrac{(x-\dfrac{3}{2})}{\dfrac{111}{4}}-\dfrac{(y+4)^2}{{\dfrac{111}{3}}}=-1\\

This is a equation of conjugated hyperbola

center(32;4)center (\frac{3}{2};-4)

semiaxes:a=1114,b=1113semi-axes : \\a=\sqrt{\frac{111}{4}}, b=\sqrt{\frac{111}{3}}


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