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The answer is in a very good situation.

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**Additional info for Student's Solutions Manual to accompany Calculus: Early Transcendental Functions**

**Example text**

A point P = (x, y) is at a distance x 2 + (y − k)2 from F and y + k from f . The equation of the curve is thus given by x 2 + (y − k)2 = y + k that is, squaring both sides x 2 + (y − k)2 = (y + k)2 . 13 The Parabola 35 Fig. 26 This reduces to x 2 − 2yk = 2yk that is y= x2 . 4k Conversely, given an equation as in the statement, the conclusion follows at once 1 1 by forcing a = 4k , that is by choosing k = 4a . This time, we notice that the origin (0, 0) is a point of the parabola with equation y = ax 2 .

4. Determine the type of the conic Γ . Determine its “metric elements” with respect to the basis R (length of the axis, position of the foci, eccentricity). 12 In a rectangular system of coordinates in solid space, consider the cone with equation x 2 + 2y 2 − 3z2 = 0. Determine all the planes whose intersection with the cone is a circle. 13 In a rectangular system of coordinates in solid space and for strictly positive numbers a, b, c, prove that the quadric abz = cxy is a hyperbolic paraboloid having two lines in common with the hyperboloid of one sheet x 2 y 2 z2 − + = 1.

Ax 2 + by 2 = z. Cutting by a plane z = d yields an ellipse when d > 0 and the empty set when d < 0. Cutting by the plane x = 0 yields the parabola by 2 = z in the (y, z)-plane and analogously when cutting by the plane y = 0. The surface has the shape depicted in Fig. 34 and is called an elliptic paraboloid. • ax 2 − by 2 = z. Cutting by a plane z = d always yields a hyperbola; the foci are in the direction of the x-axis when d > 0 and in the direction of the y-axis when d < 0. Cutting by the plane z = 0 yields √ √ √ √ ( ax + by)( ax − by) = 0 42 1 The Birth of Analytic Geometry Fig.