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Math  /  Algebra

QuestionThe ellipsoid 2x2+5y2+z2=68552/4412 x^{2}+5 y^{2}+z^{2}=68552 / 441 and sphere x2+y2+z2−3x−4y−4z−526/441=0x^{2}+y^{2}+z^{2}-3 x-4 y-4 z-526 / 441=0 are tangent to each other at a point P(a,b,c)P(a, b, c) where a=2.a=2 .
Find the remaining coordinate values. - b=b= absin⁡(a)∂∂xfa^{b} \quad \sin (a) \quad \frac{\partial}{\partial x} f : □\square □\square □\square □\square ? - c=c= absin⁡(a)∂∂xfa^{b} \quad \sin (a) \quad \frac{\partial}{\partial x} f □\square

Studdy Solution

STEP 1

1. The ellipsoid and sphere are tangent at point P(a,b,c) P(a, b, c) .
2. The coordinates of the tangent point must satisfy both the ellipsoid and sphere equations.
3. The gradient vectors of the ellipsoid and sphere at the tangent point are parallel.

STEP 2

1. Substitute a=2 a = 2 into both equations to find conditions for b b and c c .
2. Use the condition for tangency (parallel gradients) to find a relationship between b b and c c .
3. Solve the system of equations to find b b and c c .

STEP 3

Substitute a=2 a = 2 into the ellipsoid equation:
2(2)2+5b2+c2=68552441 2(2)^2 + 5b^2 + c^2 = \frac{68552}{441}
Calculate 2(2)2 2(2)^2 :
8+5b2+c2=68552441 8 + 5b^2 + c^2 = \frac{68552}{441}

STEP 4

Substitute a=2 a = 2 into the sphere equation:
(2)2+b2+c2−3(2)−4b−4c−526441=0 (2)^2 + b^2 + c^2 - 3(2) - 4b - 4c - \frac{526}{441} = 0
Calculate (2)2−3(2) (2)^2 - 3(2) :
4+b2+c2−6−4b−4c=526441 4 + b^2 + c^2 - 6 - 4b - 4c = \frac{526}{441}
Simplify:
b2+c2−4b−4c−2=526441 b^2 + c^2 - 4b - 4c - 2 = \frac{526}{441}

STEP 5

Calculate the gradient of the ellipsoid at point (2,b,c) (2, b, c) :
∇f=(4x,10y,2z) \nabla f = (4x, 10y, 2z) ∇f=(8,10b,2c) \nabla f = (8, 10b, 2c)
Calculate the gradient of the sphere at point (2,b,c) (2, b, c) :
∇g=(2x−3,2y−4,2z−4) \nabla g = (2x - 3, 2y - 4, 2z - 4) ∇g=(4−3,2b−4,2c−4) \nabla g = (4 - 3, 2b - 4, 2c - 4) ∇g=(1,2b−4,2c−4) \nabla g = (1, 2b - 4, 2c - 4)
Set the gradients parallel:
(8,10b,2c)=k(1,2b−4,2c−4) (8, 10b, 2c) = k(1, 2b - 4, 2c - 4)

STEP 6

From the parallel condition, equate the components:
8=k⋅1 8 = k \cdot 1 10b=k(2b−4) 10b = k(2b - 4) 2c=k(2c−4) 2c = k(2c - 4)
Solve for k k from the first equation:
k=8 k = 8
Substitute k=8 k = 8 into the other equations:
10b=8(2b−4) 10b = 8(2b - 4) 2c=8(2c−4) 2c = 8(2c - 4)
Simplify:
10b=16b−32 10b = 16b - 32 2c=16c−32 2c = 16c - 32

STEP 7

Solve the simplified equations:
For b b :
10b=16b−32 10b = 16b - 32 32=6b 32 = 6b b=326=163 b = \frac{32}{6} = \frac{16}{3}
For c c :
2c=16c−32 2c = 16c - 32 32=14c 32 = 14c c=3214=167 c = \frac{32}{14} = \frac{16}{7}
The remaining coordinates are:
b=163,c=167 b = \frac{16}{3}, \quad c = \frac{16}{7}

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