STEP Support Programme. STEP III Coordinate Geometry Solutions

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Grant Baldwin
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1 STEP Support Programme STEP III Coordinate Geometry Solutions These are not fully worked solutions you need to fill in some gaps. It is a good idea to look at the Hints document before this one. 1 First, we need a nice clear diagram. This one includes the circle mentioned in the second part: Note that the lines ST and UV pass through the origin, as S and T both lie on the line y = mx and U and V both lie on the line y = nx. Gradient of SV : ms nv s v Equation of SV : y ms = ms nv s v (x s Setting y = 0 and rearranging to find x gives p = (m nsv ms nv as required. Similarly, q = (m nut mt nu. S and T are the points of intersection of y = mx with x + (y c = r. Solving simultaneously, x + (mx c = r x + m x mxc + c r = 0 x mc 1 + m x + c r 1 + m = 0 s and t are the roots of this equation. (x s(x t = x (s + tx + st, so s + t = and st = c r. 1+m As U and V are the points of intersection of the line y = nx with the circle we have u + v = nc and uv = c r. 1+n 1+n mc 1+m STEP III Coord Geom Solutions 1

2 (m nsv (m nut p + q = + ms nv mt nu (m n = ((mt nusv + (ms nvut (ms nv(mt nu (m n = (mvst nsuv + must ntuv (ms nv(mt nu (m n = (m(u + vst n(s + tuv (ms nv(mt nu Substituting for s + t, u + v, st and uv gives p + q = 0 as required. STEP III Coord Geom Solutions

3 Start with a diagram: For any n, OM n sin α = r n. Note that OM n r n r n+1 = OM n+1. Substituting for OM n+1 and OM n and rearranging gives r n+1 r n = 1 sin α 1 + sin α as required. Area of C 0 = πr 0 ( 1 sin α Area of C 1 = πr1 = πr0 1 + sin α Continuing gives us a geometric series for the sum of the areas of the circles, with first term and common ratio πr 0 ( 1 sin α 1 + sin α ( 1 sin α 1 + sin α Using the formula for the sum to infinity of a geometric series, adding in the term for the semicircle, and rearranging, gives as required. S = 1 + sin α 4 sin α πr 0. STEP III Coord Geom Solutions 3

4 Thus the ratio of S to T is given by: Area of triangle = OM 0 M 0 B r 0 T = r 0 sin α cos α S T = 1 + sin α π sin α cos α 4 sin α = π 4 cos α(1 + sin α Differentiating with respect to α gives ( d S = π ( cos α sin α cos α sin α(1 + sin α. dα T 4 Setting this equal to zero we find the maximum occurs when cos α = gives S T at the maximum to be π > 3 = 16 4 > 16 5 = and substituting STEP III Coord Geom Solutions 4

5 3 As always, a diagram helps to set the scene: Differentiating a + y b = 1. with respect to x and rearranging gives x dy dx = xb a y So at P, the gradient is b cos θ a sin θ So the gradient of ON is a sin θ b cos θ So ON has equation y = a sin θ b cos θ x Next, let us find the equation of the line joining S and P. It has gradient through ( ea, 0 so its equation is given by y = b sin θ (x + ea ea + a cos θ We want to find the y coordinate at the intersection of SP and ON. equations of the lines to give x in terms of y, and equating, gives: ( ( ea + a cos θ b cos θ y ea = y b sin θ a sin θ ( ea + a cos θ = y b cos θ = ea b sin θ a sin θ ( a (e + cos θ b cos θ = y = ea sin θ ab = y(a e + (a b cos θ = ea b sin θ = y(a e + (a (a (1 e cos θ = ea b sin θ = y(a e(1 + e cos θ = ea b sin θ = y = b sin θ 1 + e cos θ b sin θ ea+a cos θ and passes Rearranging the STEP III Coord Geom Solutions 5

6 As the point T lies on ON, we can substitute into the equation for ON to get the x coordinate of T: b sin θ 1 + e cos θ = a sin θ b cos θ x so x = b cos θ a + ae cos θ The circle with centre S and radius a is the set of points (x, y satisfying (x + ea + y = a. For convenience, let = 1 + e cos θ 1. Substituting our values for x and y at the point T on the LHS gives: ( b cos θ a = b4 cos θ a + ea + + eb cos θ = (1 e b cos θ ( b sin θ + e a + b sin θ + eb cos θ + e a + b sin θ = b cos θ b e cos θ + eb cos θ + e a + b sin θ = b (1 e cos θ + eb cos θ + e a = b (1 e cos θ(1 + e cos θ + eb cos θ + e a = b (1 e cos θ + eb cos θ + e a = b (1 + e cos θ + e a = b + e a = a e a + e a = a since b a = 1 e Thus, the coordinates of T satisfy (x + ea + y = a and so T lies on the circle. 1 It is important to bear in mind that is not constant. STEP III Coord Geom Solutions 6

7 4 Note that r = x + y, and tan θ = y x. Differentiating the second expression with respect to t gives: sec θ dθ = 1 ( x y dx Hence, Then, as required. 1 dθ = y = 1 x + y x 1 x ( y dx ( y dx r dθ = 1 (x + y dθ = 1 (x + y dθ = 1 (x + y 1 x + y = 1 ( y dx ( y dx (* If P = (x, y, then A = (x a cos t, y a sin t and B = (x + b cos t, y + b sin t. Using (, [A] = 1 ( X dy Y dx, where X = x a cos t and Y = y a sin t. Note that [P ] = 1 ( y dx. [A] = 1 = 1 = 1 [ ( dy (x a cos t a cos t [ a cos tdy ( y dx = [P ] af + πa ( ] dx (y a sin t + a sin t xa cos t + a cos t y dx ] + a sin tdx ay sin t + a sin t a ( cos t dy + x cos t sin tdx + y sin t + a (cos t + sin t STEP III Coord Geom Solutions 7

8 B has coordinates as A but with a = b, so [B] = [P ] + bf + πb Since [A] = [B], πa af = πb + bf so (a + bf = π(a b so f = π(a b. The desired area is [A] [P ] = πa a ( π(a b = πab as required. STEP III Coord Geom Solutions 8

worked out from first principles by parameterizing the path, etc. If however C is a A path C is a simple closed path if and only if the starting point

III.c Green s Theorem As mentioned repeatedly, if F is not a gradient field then F dr must be worked out from first principles by parameterizing the path, etc. If however is a simple closed path in the