Question: Show that the surface x^2 - 2yz + y^3 = 4 is perpendicular to any number of the family of surfaces x + 1

Show that the surface x^2 - 2yz + y^3 = 4 is

Show that the surface x^2 - 2yz + y^3 = 4 is perpendicular to any number of the family of surfaces x + 1 = (2-4a) y^2 + a z^2. at the point of intersection (1,-1, 2).

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