Answer to Question #105595 in Analytic Geometry for Vikas

Question #105595
a) Does there exist a plane targent to x
2 −2y
2 +2z
2 = 8 and which passes through
2x+3y+2z = 8, x−y+2z = 5? Justify your answer. (5)
1
Expert's answer
2020-03-17T10:48:35-0400

Given f(x,y,z)=x22y2+2z28=0f(x,y,z)=x^{2}-2y^{2}+2z^{2}-8=0 with point P(a,b,c)P(a,b,c) .


We calculate (fx,fy,fz)=(2x,4y,4z)(f_{x},f_{y},f_{z})=(2x,-4y,4z) in this point: (2a,4b,4c)(2a,-4b,4c) .


Equation of the tangent plane is 2a(xa)4b(xb)+4c(xc)=02a(x-a)-4b(x-b)+4c(x-c)=0 , ax2by+2cz=8.ax-2by+2cz=8.


The system {2x+3y+2z=8xy+2z=5\left\{\begin{array}{ll} 2x+3y+2z = 8 \\ x-y+2z=5 \end{array}\right. is a line, we choose points on the line.


The points (3,0,1)(3,0,1) and (1,1,72)(-1,1,\frac{7}{2}) of the given line must lie in this tangent plane,


then we have c=83a2c=\frac{8-3a}{2} and b=4023a4b=\frac{40-23a}{4} .


Using the equation a22b2+2c28=0a^{2}-2b^{2}+2c^{2}-8=0 , we do not find a real solution for aa .


Answer: The tangent plane does not exist.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment