(b) Find, in terms of c and p, the coordinates of Q. (4) Do a diagram showing hyperbola, P, normal, and Q

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1 Jan 2016 Recent IAL questions on parabolas and hyperbolas, "scaffolded" 6. The rectangular hyperbola H has equation xy = c 2, where c is a non-zero constant. The point P cp, c p, where p 0, lies on H. (a) Show that the normal to H at P has equation yp p 3 x = c(1 p 4 ) The normal to H at P meets H again at the point Q. (b) Find, in terms of c and p, the coordinates of Q. (4) 1. Do a diagram showing hyperbola, P, normal, and Q 2. Using formula book, find typical point on H (ct, c/t) 3. Calculate slope of tangent = dy/dx = (dy/dt)/(dx/dt) 4. Calculate slope of normal = -1/(slope of tangent) and equation of normal 5. Put t=p to find the particular normal at P 6. Plug in y=c/q t and x=cq ttt to find points (cq,c/q) t t where normal line meets H 7. Write Step 6 q-equation gives an equation as a quadratic in t, with p and functioning solve using it as fact a constant. that (q-p) Find is a the factor sum of roots of this equation. Since you know p is a root of the equation for t (the normal line must meet H at t=p), subtracting p from the sum of roots gives you the other root, which is q. 8. Plug q-value into (cq,c/q) to find Q (5) DO NOT WRITE IN THIS AREA DO NOT WRITE IN THIS AREA DO NOT WRITE IN THIS AREA 18 *P46948A01832*

2 Jan The parabola P has equation y 2 = 4ax, where a is a positive constant. The point S is the focus of P. The point B, which does not lie on the parabola, has coordinates (q, r) where q and r are positive constants and q > a. The line l passes through B and S. (a) Show that an equation of the line l is (q a) y = r(x a) The line l intersects the directrix of P at the point C. Given that the area of triangle OCS is three times the area of triangle OBS, where O is the origin, (b) show that the area of triangle OBC is 6 5 qr (5) 1. Draw diagram showing P, B, the line l, S, C, the directrix, and O. Write in what you know about B, S, and C (its x-coordinate?) 2. Use formula book to get coordinates of S and x-coord of C 3. Use GCSE method to find eq. of line thro' 2 points B and S 4. Find Triangles area of OCS triangle and OBS OBS from have half same base height x height, from with O, base so OS. the Then base area of OCS of must triangle be OBC 3 times = 4 x base area OBS. of OBS, Only so problem CS = 3 is, BS, this so is x-difference in terms of a and between r, and we C and S is want three an answer times x-difference in terms of q and between r. S and B. Use this to find q. 5. To get an equation connecting q and a, see that triangles OBS and OCS have the same base OS, so the height of OCS relative to the base OS, i.e. the vertical distance 5. Area of OBC = area of OBS + area of OCS = 4 x area of OBS. Use your from C to the x-axis, must be three times the height of OBS, or 3r. Then: knowledge of q to find area OBS = ½ (height from B) x (base=os) EITHER: Let X be the foot of the perpendicular from B to x-axis, and Y be the meeting-point of directrix and x-axis. Then triangles CYS and BXS are similar, so XS/YS = CY/BX = 3, which gives us an equation connecting q and a. OR: We now have the coordinates of C as (-a, -3r). But C is on the line l. So we can plug x = - a, y = - 3r into the equation of line l, and get an equation connecting q and a that way. (3) DO NOT WRITE IN THIS AREA DO NOT WRITE IN THIS AREA DO NOT WRITE IN THIS AREA 26 *P46948A02632*

3 Jun y Q O S P x 1. Using formula book, write in coordinates of S and x-coordinate of Q on the diagram. C Figure 1 Cy 2 = 4ax, where a is a positive constant. The point SC and the point QC. The point P lies on C where y > 0 and the line segment QP is parallel to the x-axis. PS PQ. 2. Remember every point on parabola is same distance from focus and directrix (1) Given that the point P has x coordinate 9 a, 3. Distance PQ = difference in x-coordinates between P and Q = 13. Use that to find a. (2) PSQ. (3) 4. Area PSQ = ½ base PQ x height from S to PQ. That height = y-coordinate of P. 8 *P44971RA0828*

4 Jun H xy = 36 6 The three points P 6 p, p, Q 6 6 q, q and R 6 6 r, r, where p, q and r are distinct, H. PQ is pqy + x p + q) (4) Given that PRQR, H at the point R is parallel to the line PQ. (6) 1. Draw a diagram showing H, P, Q, and R. You'll need to have Q on other "side" of hyperbola from P and RQ to have PR perp. to QR 2. Find eq. of line through two points P and Q by GCSE method 3. Simplify that equation to find the slope of the line PQ. 3. See what slope of that line is. Deduce formulas for the slope of PR and How would you calculate the slope of PR? Same way, but with r instead of q. So the the formula slope of QR. for the slope of PQ will also give you the slope of PR if you replace q in it by r. Same sort of thinking will give you slope of QR. 4. Put slope PR = -1/slope QR, and get an equation r-squared= -1/pq 5. Get slope of normal at R from dy/dx and slope normal = -1/slope tangent 6. Use step 4 to show slope of normal at R = slope of PQ *P44971RA01228*

5 Jan 2015 This question is a bit harder than the others: 14 marks! 4. Cy 2 = 12x P(3p 2, 6p) lies on C, where p 0 CP is y + px = 6p + 3p 3 (5) CQ. Given that p S Q, (5) PQS. (4) 1. Draw a diagram showing parabola, P, S, X, Q, and what you know about their coordinates (if X is the point where the normal crosses the x-axis) 2. Find eq. of normal at P 2. by Find usual dy/dx method at general point (3t^2, 6t) by calculating dy/dt and dx/dt. Calculate slope of normal m = -1/(dy/dx). Get equation of normal by (y-6t)=m(x-3t^2). Put t=p and rearrange equation. 3. Plug p=2 into eq. of normal, and use method you learned in class to To find find Q the = (3q-squared, points where 6q) that = normal other line point crosses besides parabola, P where plug normal the general crosses parabola point x=3t^2, y=6t into that equation of the normal. You now have an equation for t. One solution must be t=2. The other must be t=q, if the point Q is described by x=3q^2, y=6q. So find q, then Q. 4. Find X by putting y=0 and p=2 into the equation of the normal 5. Area PQS = area PXS and + area SXQ 8 *P45876RA0828*

6 June The hyperbola H has cartesian equation xy = 16 The parabola P has parametric equations x = 8t 2, y = 16t. (a) Find, using algebra, the coordinates of the point A where H meets P. (3) Another point B (8, 2) lies on the hyperbola H. (b) Find the equation of the normal to H at the point (8, 2), giving your answer in the form y = mx + c, where m and c are constants. (5) (c) Find the coordinates of the points where this normal at B meets the parabola P. (6) 1. Draw a diagram with the hyperbola, the parabola, the points A and B, and the normal at B. 2. Find A by substituting general point on parabola, (8t-squared, 16t) into the hyperbola equation xy=16 and solving for t 3. Find the equation of the normal to hyperbola at B as usual 4. Plug in the typical point on the parabola into the equation of that normal - x=8t-squared, y=16t - to get an equation in t and find the two t-values where that normal crosses the parabola. 24 *P44965A02432*

7 Answers to recent IAL questions Q.6 Jan Q is ( cp 3, cp 3 ) Q.8 Jan q=5a/3 Area of triangle OBC is 6qr/5. Q.4 June (a) length of PQ = 13 (b) value of a is 4 (c) area of triangle PSQ = 78 Q.6 June Slope of tangent at R is 1/r 2, so slope of normal at R is r 2. PR is perpendicular to QR means 1/pr 1/qr = 1, so: 1/pq = slope of PQ = r 2. Q.4 Jan (b) coordinates of Q, where normal crosses the parabola again, are: (27, 18) (c) area of triangle PQS is 225 Q.8 June (a) coordinates of the point A where H meets P are (2,8) (b) equation of the normal to H at the point (8, 2) is y = 4x 30 (c) coordinates of the points where this normal at B meets the parabola P are: (12½, 20), (4½, 12)

Further Pure Mathematics F1

Write your name here Surname Other names Pearson Edexcel International Advanced Level Centre Number Further Pure Mathematics F1 Advanced/Advanced Subsidiary Wednesday 13 January 2016 Afternoon Time: 1