Find the points on the surface y2=9+xz that are closest to the origin.

Question

Clara Clark · Accepted Answer

Let (x,y,z) be a point on the surface of $y^{2} = 9 + x z$ . The distance of this point from the origin is given by:
$d = \sqrt{{(x - 0)}^{2} + {(y - 0)}^{2} + {(z - 0)}^{2}} = \sqrt{x^{2} + y^{2} + z^{2}}$
To minimise the distance, it is simpler to minimise the function $d^{2}$ .
$d^{2} = x^{2} + y^{2} + z^{2};$
$∵ y^{2} = 9 + x z$
$d^{2} = f (x, z) = x^{2} + 9 + x z + z^{2}$
$f_{x} = 2 x + z; f_{z} = x + 2 z$
Now when we solve $f_{x} = 0 and f_{z} = 0$ , we get only once critical point (0,0).
$f_{x x} = 2; f_{x z} = 1; f_{z z} = 2$
$D = f_{x x} f_{z z} - f_{x z}^{2}$
$D (0, 0) = 3 - 1 = 3 > 0; f_{x x} > 0$
Hence the point (0,0) is a point of minima using the second derivatives test.
After substituting we get $y^{2} = 9 \Rightarrow y = \pm 3$ . So the 2 points that are closest to the surface are (0,3,0) and (0, -3, 0).
Result:
(0,3,0) and (0, -3, 0)

Find the points on the surface y^{2}=9+xz that are closest

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