STEP I 2002 question 1 solution

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STEP I 2002 question 1 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 2002 question 1 solution

$(x + 2)^{2} + 2y^{2} = 18$

$9(x - 1)^{2} + 16y^{2} = 25$ .

For there to be an intersection these equations must be true simultaneously, and therefore, at the same time they must have the same y-value:

$y^{2} = \frac{18 - (x + 2)^{2}}{2}$

Substituting this:

$9(x - 1)^{2} + 16y^{2} = 9(x - 1)^{2} + 144 - 8(x + 2)^{2} = 25$

Hence:

$9x^{2} - 18x + 9 + 144 - 8x^{2} - 32x - 32 = 25$

$x^{2} - 50x + 121 = 25$

Hence:

$x^{2} - 50x + 96 = 0 = (x - 48)(x - 2)$ .

Now, calculate the corresponding y-values:

$x = 48 \implies (48 + 2)^{2} + 2y^{2} = 18 \implies y^{2} = -1241$ .

There is no defined y-value at for this "x" (the other equation can be used to corroborate this).

$x = 2 \implies (2 + 2)^{2} + 2y^{2} = 18 \implies y = \pm 1$ .

Hence, the circle must pass through $(2, \ \pm 1)$ .

Take the form given:

$x^{2} - 2ax + y^{2} = 5 - 4a$ .

To show that a circle which passes through the aforementioned points can be written in this style check to see what the y-value is for the given x-value:

$x = 2 \implies x^{2} - 2ax + y^{2} = 4 - 4a + y^{2} = 5 - 4a \implies y = \pm 1$ .