 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2016/8/solution

Split the mixed quadratic terms evenly across a real symmetric matrix:

A = 1 - 2 - 2 - 2 - 4 2 - 2 21, Q_{1} = r^{T} A r .

det (λ I - A) = (λ + 1) (λ + 5) (λ - 4) = λ^{3} + 2 λ^{2} - 19 λ - 20,

so $c = 2$ . Choose the orthonormal eigenvectors

u_{1} = \frac{( 1 , 0 , 1 ) ^{T}}{2}, u_{2} = \frac{( - 1 , - 4 , 1 ) ^{T}}{3 2}, u_{3} = \frac{( - 2 , 1 , 2 ) ^{T}}{3}, P = (u_{1} u_{2} u_{3}) .

r = P R

gives

Q_{1} = - X^{2} - 5 Y^{2} + 4 Z^{2}, Q_{2} = X^{2} + Y^{2} + Z^{2}, β = - 5, γ = 4.

Indeed

P^{T} P = I

and

P^{T} A P = diag (- 1, - 5, 4)

; preserving

Q_{2}

is precisely the length-preserving property of an orthogonal matrix.

The requested plane has normal

(2, 1, - 2)

. An orthonormal basis within it is

e_{1} = \frac{( 1 , 0 , 1 )}{2}, e_{2} = \frac{( 1 , - 4 , - 1 )}{3 2} .

Write a point of that plane as

r = u e_{1} + w e_{2}

. Restricting the quadratic form gives

Q_{1} = - u^{2} - (13/9) w^{2}

, so its section with

Q_{1} = - 5

\frac{u ^{2}}{5} + \frac{w ^{2}}{45/13} = 1.

It is an ellipse centred at the origin, with semi-axes

5

along

e_{1}

and

45/13

along

e_{2}

. Normalizing the in-plane basis is essential: coefficients in nonorthonormal coordinates would not directly give physical lengths.

\frac{X ^{2}}{5} + Y^{2} - \frac{Z ^{2}}{5/4} = 1.

It is connected, extends indefinitely in both directions along its axis

u_{3}

, and has an elliptical waist at

Z = 0

5

and

1

. At larger

∣ Z ∣

its transverse sections widen; its asymptotic cone is

4 Z^{2} = X^{2} + 5 Y^{2}

. The specified plane is oblique to that axis, rather than the waist plane.

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