Question

Which of the following is the equation of the plane that is tangent to the surface 222 + y² - zz-32= -1 at the point P (1,-1,1)? 3 x + 2y - 42 = -3 5x – 2y - 42 = 3 3x - 2y - 32 = 2 3x - 2y - 42 = 1 52 – 2 – 3z = 4 00 Mark This Question

Answer 1

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Ane Given surface is 2x + y - x2 -32=-) » 2x2+ y +) = x2+3Z 2x++ y2 +) = 2 (x+3) = 2x² + y² + - Z x + 3 =) 2 = 2x+y + (X+3) We know that equation of tangent blane to surface z=f(x,y) at point (xo, Yo, zo) is given by 2- 2. = fe(40, 7o) (x -X.) + fy(70,90) (9-7.)-

we Compairing 2=fa,y) with 6, get fa,y)=2x² + y² +) (x+3) We need equation at the point 3 (1,-1, 1), 30 (7,9 2) = (1-1) (so Xo = 1, y =-1, 20=)) Now fx (4.7) = a (tax,y)] by (3) Jul (x+3) (x+3)(2x+yº+1) - (2x + y* +) y (C+3) (x+3)2 da ge+y+1

here h(x)=x+3 (x)+d (3) JX Jel hao as a g(x,y) h(x) Ja (914,99) – 94,9)x(her)) [6(x)] g(x,y) = 2x+y +, ***>>>>() - (92"+y+" ** (x+3) (2(21) +0 to] - (2x+y +1) ((1) +o] (a + 3) as (xn) = mxn), Constant) - o) (919))=0 (X+3) (4x) - (2x + y²+1) (x+3) 2(2x) toto - (x+3) 2

- کو دو- روا رہا - (3+3) - 2 ا- دوا+ - دو مرد + 3) (- د ) + و - دو = (9) 4 ( ورد+1) Cet x= 1, y=-) (- (1) 3/-را-) ) = (1- ا د (+3) ا- 9+ ( - 2 ه کرد) 3 و |

- fx (1,-1) = 3 9 4 sy he (x+3) - Now fy (x,y) = 2 (fray, 7 fy(3,1)=(2x+y++) (by *) (20) 33(2x++y+1) (2x°) + (y) +2 (0) as & (g(x)) = 0 (*c+3) sy a lyn)= myn- sy on Iconstan #)= (x+3) he ] he of het a P he

키 fg12,9)= 2y (1+3) at려, 며 2 10,1)= 2 (-1) & 음 그 +3 5) 1) - ) Now but (xo, J., 20) =(1-1, 1) in > 2 1 = f(,)) (x) + fy (,) (-(-1)) 22 7 -) = () (X-)) + (J) (+1) (O & ⑤) 쮸-류블 키니레) = 31 -3-2y -2 키 러= 32 니 2.

=) 42-4=30-2-5 =) 5-4= = 3x - 24-42 =) 1 = 3x-24-42 3- 24 – 42 = 1 (fourth option is correct.