STEP III 1997 question 8 solution

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STEP III 1997 question 8 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP 1997 Solutions > STEP III 1997 question 8 solution

(i)

If we just carry out the multiplication, we'll find that the off-diagonals are the same. If we equate either to zero, we get the following equation:

$-a \sin (\alpha) \cos (\alpha) + b \cos ^2(\alpha) - b \sin ^2(\alpha) + c \cos(\alpha) \sin (\alpha) = 0$

$\Longrightarrow (c-a) \sin (\alpha) \cos (\alpha) + b( \cos ^2(\alpha) - \sin ^2(\alpha)) = 0$

$\Longrightarrow \frac 12 (c-a) \sin (2\alpha) + b \cos (2\alpha) = 0$

$\displaystyle \Longrightarrow \tan (2\alpha) = \frac{2b}{a-c}$

$\Longrightarrow \alpha = \frac 12 \arctan \left(\frac{2b}{a-c}\right)$

(ii)

Let's expand the equation of the ellipse:

$x^2 + y^2 + 4xy \cot (2\theta) + 4x^2 \cot ^2(2\theta) = 1$

And also, let's expand the matrix equation:

ax^2 + 2bxy + cy^2 = 1

Now let's equate coefficients:

$a = 1 + 4 \cot ^2(2\theta)$

$b = 2 \cot (2\theta)$

c = 1

So we have what A looks like in this case.

If it's diagonal, then:

$\displaystyle \alpha = \frac 12 \arctan \left(\frac{2b}{a-c}\right)$

$\displaystyle= \frac 12 \arctan \left(\frac{4 \cot (2\theta)}{4 \cot ^2(2\theta)}\right)$

$\displaystyle = \frac 12 \arctan \left(\frac{1}{ \cot (2\theta)}\right)$

$\displaystyle = \frac 12 \arctan ( \tan (2\theta))$

$\displaystyle = \theta$ , as required.

Let's go back to the $R_{-\alpha} A R_\alpha$ we found in the first part of the question. Here we need to replace a,b,c with our new values, and $\alpha$ with $\theta$ . We find, upon simplification, that the non-diagonal elements are (respecitvely, top left and bottom right):

$\cot ^2(\theta)$

$\tan ^2(\theta)$

(iii) What $R_{-\alpha} A R_\alpha$ does is shift the xy-axes into x'y'-axes so that they align with the major and minor axes of the ellipse. Let's write down the new equation of the ellipse (using the matrix equation in part (ii), but using $R_{-\alpha} A R_\alpha$ instead of ; and (x', y') instead of (x, y) , to emphasise the coordinate switch):

$(x')^2 \cot ^2(\theta) + (y')^2 \tan ^2(\theta) = 1$

Here, the minor axis has length $\tan (\theta)$ and the major axis has length $\cot (\theta)$ . Done.

Solution by dvs

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