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IgorC [24]
3 years ago
5

At what point on the paraboloid y = x2 z2 is the tangent plane parallel to the plane 3x 2y 5z = 8?

Physics
1 answer:
salantis [7]3 years ago
6 0
To find the tangent plane to the surface f(x,y,z)=0 at a point (X,Y,Z) we use the following method: 

<span>Calculate grad f = (f_x, f_y, f_z). The normal vector to the surface at the point (X,Y,Z) is grad f(X,Y,Z). The equation of a plane with normal vector n which passes through the point p is (r-p).n=0, where r=(x,y,z) is the position vector. So the equation of the tangent plane to the surface through the point (X,Y,Z) is ((x,y,z)-(X,Y,Z)).grad f(X,Y,Z)=0. </span>

<span>Now in your case we have f(x,y,z)=y-x^2-z^2, so grad f=(-2x,1,-2z), and the equation of the tangent plane at the point (X,Y,Z) is </span>

<span>((x,y,z)-(X,Y,Z)).(-2X,1,-2Z)=0, </span>

<span>that is </span>

<span>-2X(x-X)+1(y-Y)-2Z(z-Z)=0, </span>

<span>i.e. </span>

<span>-2Xx+y-2Zz = -2X^2+Y-2Z^2. (1) </span>

<span>Now compare this equation with the plane </span>

<span>x + 2y + 3z = 1. (2) </span>

<span>The two planes a_1x+b_1y+c_1z=d_1, a_2x+b_2y+c_2z=d_2 are parallel when (a_1,b_1,c_1) is a multiple of (a_2,b_2,c_2). So the two planes (1),(2) are parallel when (-2X,1,-2Z) is a multiple of (1,2,3), and we have </span>

<span>(-2X,1,-2Z)=1/2(1,2,3) </span>

<span>for X=-1/4 and Z=-3/4. On the paraboloid the corresponding y coordinate is Y=X^2+Z^2=1^4+9^4=5/2. </span>

<span>So the tangent plane to the given paraboloid at the point (-1/4,5/2,-3/4) is parallel to the given plane.</span>
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