Question

Find the equation of the plane tangent to the following surface at the given points. x2 + y2 - 2? + 5 = 0; (4,2,5) and (-2,-4,5) The equation of the tangent plane at (4,2,5) is = 0.

Answer 1

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a surface f(x, y, z)=( - equation at a of tangent plane to point (10, Yo, 7o) is (4-co). of levy, z) = 0 that is z (xo, Yo, zo (xo, Yo, Zo) (xo, Yo iZo) 0714,2,5) dy 114,2,5) given suttare f(x,y,z) = c is x?+y?-z' +5 -0 and point (x0, 4o, Zo) is (4,2,5 ) = (xo, Yo, zu) Now to get b (x'+y?2+5) = 2x2 = 2x4 = 8 of - dp (xty?-z’+5) = 2y = = 2x2 = 4 of = (x+y?-z?+5) = -22 = -2X5= -40 Now equation of tangent plane to suetace x²+y?z? +5 = 0 at (4, 2,5) is (2-4)(8) + (4-2)(4) + (7-5) (-10) = o Cusing on 0] 21(4,2,5) - 8x+4y -luz + 10 =0 Now at point (aro, Yo Zo) = (-2,-4,5) c) = 2X(-2) = -4, I = 2(-4) = -8 ) og (-2-4,5) (-2,-4,5) = -2(5) = -10 hence equation of tangent plane to sustace x² + y² z?ts=0 at (-2, -4,5) is (use equation o) (ac+2)(-4) +(4+4)(-8) + (2-5) (-0) = 0 + - 431 – 84 – loz +10=0