Quadric Surfaces — Question 3

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Question 3

For 9x2−4y2+9z2+36x+8y−18z+45=09x^2-4y^2+9z^2+36x+8y-18z+45=0, find the standard form, classify the surface, and determine the vertices and which coordinate planes fail to intersect it.

Original worksheet page 1: question and worked solution for 6-4-003
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Question 3 – Solution

Strategy Complete squares; a single negative term with right side −1-1 is more clearly read after multiplying by −1-1.

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Standard form Completing squares gives 9(x+2)2−4(y−1)2+9(z−1)2=−4,9(x+2)^2-4(y-1)^2+9(z-1)^2=-4, hence (y−1)21−(x+2)24/9−(z−1)24/9=1.\boxed{\frac{(y-1)^2}{1}-\frac{(x+2)^2}{4/9}-\frac{(z-1)^2}{4/9}=1}. This is a hyperboloid of two sheets with axis parallel to yy.

Vertices and coordinate traces The vertices are (−2,0,1),(−2,2,1)\boxed{(-2,0,1),(-2,2,1)}. All three coordinate planes intersect the surface: y=0y=0 contains the first vertex, while x=0x=0 and z=0z=0 give hyperbolas. The plane y=1y=1 does not intersect the surface, but it is not a coordinate plane.

Original worksheet page 2: question and worked solution for 6-4-003

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