Solutions 5 MT1100 From Euclid to Mandelbrot 1. The square of the distance from (x, y) to the focus is (x - ae)2 + y2. We need to show that this is e2 times the square of the distance from (x, y) to the directrix. The equation of the ellipse shows that y2 = 62 - 2x2 and so (x - ae)2 + y2 = x2 - 2aex + a2e2 + y2 = x2 - 2aex + a2+2 + b2 -2x2 = (1 - 2) x2 - 2aex + a2e2 + b2 = e2x2 - 2aex + a2 - b2 + b2 (using the formula for e twice) = e2(22-2,0 + 2) = e2 (x -a)2 . Since (x - a)- is the square of the distance from (x, y) to the directrix, this shows what we want. 2. When the ellipse is written in standard form, the directrices are the lines x = +a/e. So | NN'| = 2a/e. We also know that | FP| = e| PN | and | F'P| = e|PN'| (see the previous question). So |FP| + | F'P| = e(|PN|+ |PN']) = e|NN'] = 2a. 3. We complete the square to get rid of any x and y terms: 16x2 - 9y2 + 96x + 36y - 36 = 16(x2 + 6x) - 9(y2 - 4y) - 36 = 16(x+3)2-9(y-2)2-36-144+36 1
so we have 16(x + 3)2 - 9(y - 2)2 = 144, or (x + 3)2 9 ? (y - 2)2 16 = 1. This is a hyperbola with centre (-3,2) and semiaxes a = 3 and b = 4. See the diagram below. The asymptotes (the blue lines in the diagram below) are given by 9 (x+ 3)2 ? (y - 2)2 =0 16 which simplifies to 4(x + 3) = +3(y - 2) or the two lines 4x - 3y = - 18 and 4x+ 3y = - 6. 10 ). 4-15. ** - 10 -5 5 .10 -5 4. Here a = 7, h = 31/3 and b = 13, so ab - h2 = 91 - 27 = 64 > 0 and so we have an equation of an ellipse. Since there are no terms in x or y, the centre is at the origin. To rotate axes, we put x = x* cos ø - y* sin ø and y = x* sin ø + y* cos o. As we know that Ø = T/3, we have x= 3x* - 5 y and ? 3 ? + ? . 1 2
Substituting these gives 7 (?x *- 15 y*) +6v3(/x*???*)(vx*+?*)+ 2 13 (13x* + +y* ) - 16 = 16(x*)2 + 4(y*)2 - 16 = 0, which can be written as ( ** )2 + (y*)2 4 == 1. 3