Alright — let’s work through this step-by-step. --- ## 1. Understanding the Problem We are given three elliptic paraboloids: \[ P_1: \quad \frac{x^2}{a_1^2} + \frac{y^2}{b_1^2} = \frac{2z}{c_1} \] \[ P_2: \quad \frac{x^2}{a_2^2} + \frac{y^2}{b_2^2} = \frac{2z}{c_2} \] \[ P_3: \quad \frac{x^2}{a_3^2} + \frac{y^2}{b_3^2} = \frac{2z}{c_3} \] We must prove numerically that they have a common tangent plane if \[ \Delta = \begin{vmatrix} a_1^2 & a_2^2 & a_3^2 \\ b_1^2 & b_2^2 & b_3^2 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0. \] "Prove numerically" probably means: find a numerical example that satisfies \( \Delta = 0 \) and actually compute a common tangent plane. --- ## 2. General equation of tangent plane For an elliptic paraboloid of the form \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{2z}{c}, \] rewrite as: \[ F(x,y,z) = \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2z}{c} = 0. \] At point \( (x_0, y_0, z_0) \) on this surface, the tangent plane is: \[ \frac{2x_0}{a^2} (x - x_0) + \frac{2y_0}{b^2} (y - y_0) - \frac{2}{c} (z - z_0) = 0. \] But since \( z_0 = \frac{c}{2} \left( \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} \right) \), we can express the plane in simpler form: Let \( u = \frac{x_0}{a^2} \), \( v = \frac{y_0}{b^2} \). Then \( z_0 = \frac{c}{2} (a^2 u^2 + b^2 v^2) \) perhaps? Wait carefully: \( x_0 = a^2 u \), \( y_0 = b^2 v \), actually, no: \( u = \frac{x_0}{a^2} \), so \( x_0 = a^2 u \), yes. Then \( z_0 = \frac{c}{2} \left( \frac{a^4 u^2}{a^2} + \frac{b^4 v^2}{b^2} \right) = \frac{c}{2} \left( a^2 u^2 + b^2 v^2 \right) \). Now tangent plane equation: \[ \frac{2x_0}{a^2} (x - x_0) + \frac{2y_0}{b^2} (y - y_0) - \frac{2}{c} (z - z_0) = 0. \] Divide by 2: \[ \frac{x_0}{a^2} (x - x_0) + \frac{y_0}{b^2} (y - y_0) - \frac{1}{c} (z - z_0) = 0. \] Substitute \( x_0 = a^2 u \), \( y_0 = b^2 v \): \[ u (x - a^2 u) + v (y - b^2 v) - \frac{1}{c} \left[ z - \frac{c}{2} (a^2 u^2 + b^2 v^2) \right] = 0. \] Simplify: \[ u x + v y - (a^2 u^2 + b^2 v^2) - \frac{z}{c} + \frac12 (a^2 u^2 + b^2 v^2) = 0. \] \[ u x + v y - \frac{z}{c} - \frac12 (a^2 u^2 + b^2 v^2) = 0. \] Thus tangent plane: \[ u x + v y - \frac{z}{c} = \frac12 (a^2 u^2 + b^2 v^2). \quad\quad (1) \] Here \( (u, v) \) are parameters related to contact point \( (a^2 u, b^2 v, \frac{c}{2}(a^2 u^2 + b^2 v^2) ) \). --- ## 3. Common tangent plane condition For a common tangent plane to all three paraboloids, there must exist \( u, v \) such that the *same* plane (1) for \( (a_i, b_i, c_i) \) has possibly different right-hand side constants for each? Wait, no — if it’s the same plane, then \( u, v \) must be the same for each, and \( \frac{1}{c} \) coefficient for \( z \) differs unless \( 1/c \) is scaled… Let’s check: Plane: \( u x + v y - \frac{z}{c_i} = k_i \) with \( k_i = \frac12 (a_i^2 u^2 + b_i^2 v^2) \). But these three equations represent the same plane in \( x, y, z \) only if the coefficients of \( x, y, z \) and the constant term are proportional or equal. Coefficient of \( x \) is \( u \), coefficient of \( y \) is \( v \) for all i — same by assumption (same \( u, v \)). Coefficient of \( z \) is \( -\frac{1}{c_i} \). For them to be the same plane in \( \mathbb{R}^3 \), all \( -\frac{1}{c_i} \) must be equal. That would require \( c_1 = c_2 = c_3 \). But the problem doesn’t require \( c_i \) equal; so maybe they mean “common tangent plane” not in the sense of identical plane equation coefficients, but that the three surfaces are all tangent to the same plane at possibly different points — not “at the same point”. Yes: a common tangent plane can touch each at different points. But then “common tangent plane” means there exists a single plane \( \alpha x + \beta y + \gamma z = \delta \) that is tangent to all 3 surfaces. Let’s use a standard fact: For \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{2z}{c} \), the tangent plane at \( (x_0, y_0, z_0) \) is \[ \frac{x