 Vectors and conics

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4.9 Further exercises

Example 58

Determine the equation of the circle with centre (2, 1) and radius 3.

The equation of the circle is

(x − 2)2 + (y − 1)2 = 32,

which can be rewritten in the form

x2 + y2 − 4x − 2y − 4 = 0.

Example 59

Determine the points of intersection of the line with equation y = x + 2 and the circle in Exercise 58. At the points of intersection of the line and the circle, the following equations must be satisfied:

y = x + 2

and

x2 + y2 − 4x − 2y − 4 = 0.

Substituting the expression for y from the first equation into the second equation, we obtain

x2 + (x + 2)2 − 4x − 2(x + 2) − 4 = 0.

Multiplying this out and collecting terms, we obtain

2x2 − 2x − 4 = 0,

or

x2 − x − 2 = 0.

This can be factorised as

(x − 2)(x + 1) = 0,

so x = 2 or x = −1.

Substituting these values of x into the equation y = x + 2 of the line, we find that the corresponding values of y are y = 4 and y = 1.

Hence the points of intersection of the line and the circle are (2, 4) and (−1, 1).

Example 60

This question concerns the parabola y2 = 4ax (a > 0) with parametric equations x = at2, y = 2at and focus F. Let P and Q be points on the parabola with parameters t1 and t2, respectively. Prove that:

• (a) if PQ subtends a right angle at the vertex O of the parabola, then t1t2 = −4;

• • (b) if t1 = 2 and PQ is perpendicular to FP, then .

• (a) The gradient of OP is

• • and the gradient of OQ is

• • OP and OQ are perpendicular (in other words, PQ subtends a right angle at O) if m1m2 = −1, that is, if

• • which reduces to t1t2 = −4, as required.

• (b) Here P is the point (4a, 4a). Also, F is (a, 0), so the gradient of FP is

• • and the gradient of PQ is

• • FP and PQ are perpendicular if m1m2 = −1, that is, if

• • which reduces to , so , as required.

Example 61

This question concerns the rectangular hyperbola xy = c2 (c > 0) with parametric equations x = ct, y = c/t. Let P and Q be points on the hyperbola with parameters t1 and t2, respectively.

• (a) Determine the equation of the chord PQ.

• • (b) Determine the coordinates of the point N where PQ meets the x-axis.

• (c) Determine the midpoint M of PQ.

• (d) Prove that OM = MN, where O is the origin.

• (a) The gradient of the chord PQ is

• • so the equation of PQ is

• • that is,

• • (b) At the point N where PQ meets the x-axis, y = 0 and equation (S.19) also holds, so x = c(t1 + t2). Hence N is the point with coordinates (c(t1 + t2), 0).

• (c) The midpoint M of PQ is

• • (d) By the Distance Formula, the distance OM is

• • Also,

• • Thus OM = MN.

• Remark: An alternative way of proving part (d) is as follows: Since the point M is vertically above the midpoint of ON, the triangle OMN must be an isosceles triangle, so OM = MN.

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