Match the equation with its graph x2 4y2 9z2 1

Asked by admin @ 27/11/2021 in Mathematics viewed by 334 People

Consider the equation below. X2 + y2 − 2x − 4y − z + 5 = 0 reduce the equation to one of the standard forms. Classify the surface. Ellipsoid elliptic paraboloid hyperbolic paraboloid cone hyperboloid of one sheet hyperboloid of two sheets sketch the surface. Webassign plot webassign plot webassign plot webassign plot

Answered by admin @ 27/11/2021

Paraboloids are generated by rotating a parabola across its axes.

The equation of the paraboloid is: $\mathbf{z = (x- 1)^2 + (y - 2)^2}$
The surface is an infinite paraboloid

The equation is given as:

$\mathbf{x^2 + y^2 - 2x - 4y - z + 5 = 0}$

Rewrite as:

$\mathbf{x^2 - 2x+ y^2 - 4y - z + 5 = 0}$

Add 0 to both sides

$\mathbf{x^2 - 2x + 0+ y^2 - 4y - z + 5 = 0 + 0}$

Express 0 as the difference between the squares of half the coefficient of x

$\mathbf{x^2 - 2x + (\frac{2}{2})^2 - (\frac{2}{2})^2 + y^2 - 4y - z + 5 = 0 + (\frac{2}{2})^2 - (\frac{2}{2})^2}$

Cancel out the negative term

$\mathbf{x^2 - 2x + (\frac{2}{2})^2 + y^2 - 4y - z + 5 = 0 + (\frac{2}{2})^2 }$

Simplify

$\mathbf{x^2 - 2x + 1 + y^2 - 4y - z + 5 = 0 +1 }$

$\mathbf{x^2 - 2x + 1 + y^2 - 4y - z + 5 = 1}$

Express as squares

$\mathbf{(x- 1)^2 + y^2 - 4y - z + 5 = 1}$

Repeat the same process for y

$\mathbf{(x- 1)^2 + y^2 - 4y + (2)^2 - z + 5 = 1 + (2)^2}$

$\mathbf{(x- 1)^2 + y^2 - 4y + 4 - z + 5 = 1 + 4}$

$\mathbf{(x- 1)^2 + y^2 + (y - 2)^2 - z + 5 = 1 + 4}$

$\mathbf{(x- 1)^2 + (y - 2)^2 - z + 5 = 1 + 4}$

$\mathbf{(x- 1)^2 + (y - 2)^2 - z + 5 = 5}$

Subtract 5 from both sides

$\mathbf{(x- 1)^2 + (y - 2)^2 - z = 0}$

Make z the subject

$\mathbf{z = (x- 1)^2 + (y - 2)^2}$

The above equation is the surface of an infinite paraboloid

See attachment for the sketch of the equation.

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