(a) Show that the planes, which cut the cone ax² + by² + cz² = 0 in perpendicular generators, touch the cone (x²)/(b+c) +…

(a) Planes cutting ax² + by² + cz² = 0 in perpendicular generators

Let the plane be lx + my + nz = 0; a plane meeting the cone in two generators must pass through the vertex O. Let n̂ = (l, m, n)/√(l² + m² + n²) and choose orthonormal vectors u, v in the plane, so that (u, v, n̂) is an orthonormal frame.

In this frame the quadratic form Q = ax² + by² + cz² restricts to the plane as Au² + 2Huv + Bv² = 0. This gives the two generators, with slopes t = u/v satisfying At² + 2Ht + B = 0. They are perpendicular exactly when t₁t₂ = B/A = −1, that is A + B = 0.

The trace of Q is independent of the orthonormal frame: A + B + N = a + b + c, where N = (al² + bm² + cn²)/(l² + m² + n²) is the value of Q along the normal. Hence

A + B = 0 ⟺ (a + b + c)(l² + m² + n²) = al² + bm² + cn² ⟺ (b + c)l² + (c + a)m² + (a + b)n² = 0. (1)

Now consider T: x²/(b + c) + y²/(c + a) + z²/(a + b) = 0. The tangent plane to T at (x₁, y₁, z₁) on it is xx₁/(b + c) + yy₁/(c + a) + zz₁/(a + b) = 0. The plane lx + my + nz = 0 is therefore tangent to T if (x₁, y₁, z₁) = λ(l(b + c), m(c + a), n(a + b)) lies on T. Substituting gives λ²[(b + c)l² + (c + a)m² + (a + b)n²] = 0, which is condition (1).

Hence every plane cutting C in perpendicular generators touches T. ∎

(b) f(x, y) = |x² − y²|: f_xy(0,0) and f_yx(0,0)

f_x(0,0) = lim (|h²| − 0)/h = lim h = 0.

For k ≠ 0 and |h| < |k|, f(h, k) = k² − h² and f(0, k) = k², so f_x(0, k) = lim (−h²)/h = 0.

Hence f_xy(0,0) = lim [f_x(0,k) − f_x(0,0)]/k = 0.

Similarly f_y(0,0) = 0, and for h ≠ 0 and |k| < |h|, f(h, k) = h² − k² and f(h, 0) = h², so f_y(h, 0) = lim (−k²)/k = 0. Hence f_yx(0,0) = lim [f_y(h,0) − f_y(0,0)]/h = 0.

Therefore f_xy(0,0) = f_yx(0,0) = 0.

(c) The subspace S = {(x, 2y, 3x)}

Every element is x(1, 0, 3) + y(0, 2, 0). The zero vector (x = y = 0) is in S. If (x, 2y, 3x) and (x′, 2y′, 3x′) are in S, so are their sum (x + x′, 2(y + y′), 3(x + x′)) and every scalar multiple (tx, 2(ty), 3(tx)). So S is a subspace of R³.

The vectors (1, 0, 3) and (0, 2, 0) span S and are linearly independent, so {(1, 0, 3), (0, 2, 0)} is a basis. A second basis is {(1, 2, 3), (0, 2, 0)}, since (1, 2, 3) is in S and the two vectors are independent.

dim S = 2.