Question

Prove that the paraboloids
x^2/a1^2 + y^2/b1^2 = 2z/c1
x^2/a2^2 + y^2/b2^2 = 2z/c2
x^2/a3^2 + y^2/c3^2 = 2z/c3 have a common teangent plane if
|a1^2 a2^2 a3^2|
|b1^2 b2^2 b3^2|
|c1^2 c2^2 c3^2|

Answer 1

Answer on Question #78594 – Math – Analytic Geometry Question

Prove that the paraboloids

$$\frac {x ^ {2}}{a _ {1} ^ {2}} + \frac {y ^ {2}}{b _ {1} ^ {2}} = \frac {2 z}{c _ {1}}$$ $$\frac {x ^ {2}}{a _ {2} ^ {2}} + \frac {y ^ {2}}{b _ {2} ^ {2}} = \frac {2 z}{c _ {2}}$$ $$\frac {x ^ {2}}{a _ {3} ^ {2}} + \frac {y ^ {2}}{b _ {3} ^ {2}} = \frac {2 z}{c _ {3}}$$

have a common tangent plane if

$$\left| \begin{array}{c c c} a _ {1} ^ {2} & a _ {2} ^ {2} & a _ {3} ^ {2} \\ b _ {1} ^ {2} & b _ {2} ^ {2} & b _ {3} ^ {2} \\ c _ {1} ^ {2} & c _ {2} ^ {2} & c _ {3} ^ {2} \end{array} \right| = 0$$

Solution

A normal vector to the surface $f(x,y,z) = c$ at $(x_0, y_0, z_0)$ is given by

$$\nabla f (x _ {0}, y _ {0}, z _ {0})$$

Then the normal vector $n_i$ to the paraboloid $P_i \colon \frac{x^2}{a_i^2} + \frac{y^2}{b_i^2} - \frac{2z}{c_i} = 0$ is given by

$$\overrightarrow {n _ {i}} = \left(\frac {x _ {0} {} ^ {2}}{a _ {i} ^ {2}}, \frac {y _ {0} {} ^ {2}}{b _ {i} ^ {2}}, - \frac {2 z _ {0}}{c _ {i}}\right)$$

If there exists a common tangent plane then

$$\left\{ \begin{array}{l} \overrightarrow {n _ {1}} = \lambda_ {2} \overrightarrow {n _ {2}} \\ \overrightarrow {n _ {1}} = \lambda_ {3} \overrightarrow {n _ {3}} \end{array} \right.$$

We have that

$$\left\{ \begin{array}{l} \frac {x _ {0} {} ^ {2}}{a _ {1} ^ {2}} = \lambda_ {2} \frac {x _ {0} {} ^ {2}}{a _ {2} ^ {2}} \\ \frac {y _ {0} {} ^ {2}}{b _ {1} ^ {2}} = \lambda_ {2} \frac {y _ {0} {} ^ {2}}{b _ {2} ^ {2}} \\ \frac {2 z _ {0}}{c _ {1}} = \lambda_ {2} \frac {2 z _ {0}}{c _ {2}} \end{array} \right.$$ $$\left\{ \begin{array}{l} \frac {x _ {0} {} ^ {2}}{a _ {1} ^ {2}} = \lambda_ {3} \frac {x _ {0} {} ^ {2}}{a _ {3} ^ {2}} \\ \frac {y _ {0} {} ^ {2}}{b _ {1} ^ {2}} = \lambda_ {3} \frac {y _ {0} {} ^ {2}}{b _ {3} ^ {2}} \\ \frac {2 z _ {0}}{c _ {1}} = \lambda_ {3} \frac {2 z _ {0}}{c _ {3}} \end{array} \right.$$

This means that the columns in the matrix

$$\left( \begin{array}{c c c} a _ {1} ^ {2} & a _ {2} ^ {2} & a _ {3} ^ {2} \\ b _ {1} ^ {2} & b _ {2} ^ {2} & b _ {3} ^ {2} \\ c _ {1} ^ {2} & c _ {2} ^ {2} & c _ {3} ^ {2} \end{array} \right)$$

are linearly dependent.

Therefore,

$$\left| \begin{array}{c c c} a _ {1} ^ {2} & a _ {2} ^ {2} & a _ {3} ^ {2} \\ b _ {1} ^ {2} & b _ {2} ^ {2} & b _ {3} ^ {2} \\ c _ {1} ^ {2} & c _ {2} ^ {2} & c _ {3} ^ {2} \end{array} \right| = 0$$

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