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STEP III 1990 question 6 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 1990 question 6 solution

\displaystyle \left(\begin{array}{c c}X\\Y \end{array}\right) = \frac{2}{5}\left(\begin{array}{c c}9 & -2 \\-2 & 6 \end{array}\right) \left(\begin{array}{c c}x\\y \end{array}\right)

So applying the transformations to the vectors (1, 2) and (2, -1), we get (2, 4) and (8, -4), both of which are scalar multiples of the original vectors.

The matrix transforms the general point (x, y) to (X, Y) = \frac{2}{5}(9x-2y, 6y-2x)

5X = 18x - 4y

5Y = 12y - 4x

25X^{2} = 324x^{2} - 144xy + 16y^{2}

25XY = 232xy - 72x^{2} - 48y^{2}

25Y^{2} = 144y^{2} - 96xy + 16x^{2}

Therefore the function x^{2} + y^{2} = 1 is transformed to:

8X^{2} + 12XY + 17Y^{2} = \frac{2000x^{2} + 2000y^{2}}{25} = 80

As required.

The area is equal to the original area multiplied by the determinant of the matrix transformation. The original area is \pi; the determinant is 8. Therefore the area is 8\pi.

To find the maximum value of X we find \frac{dX}{dY}

Differentiating implicitly, we get \frac{dX}{dY} = - \frac{6x + 17y}{8x + 6y}

There is a stationary point when Y = \frac{-6x}{17}

Treating the curves equation as a quadratic in Y, we get Y = \frac{-6x \pm \sqrt{1360 - 100X^{2}}}{17}

At the stationary point, 1360 - 100X^{2} = 0 \Rightarrow X^{2} = \frac{68}{5}

\Rightarrow X = 2\sqrt{\frac{17}{5}}

The point (0.6, 0.8) is transformed to the point (\frac{38}{25}, \frac{36}{25}).

At this point, \frac{dY}{dX} = -\frac{13}{21}

The equation of the tangent at this point is therefore 21Y + 13X = 50.

Solution by Dystopia.

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