8. ABC is an acute-angled triangle with circumcentre P. If AD is the altitude to BC, prove that the bisector of angle A bisects <DAP. [5] See attached diagrams 2. Either: Let bisector of Angle A meet the circumcircle at X, √ Let AP meet the circumcircle at E. Join BE. <ACD = <AEB since <s at circumf., same arc √ Line APE is a diameter so <ABE = 90 √ Triangles ABE and ADC have two pairs of equal angles, so are similar, Hence <BAE = <DAC √ So <EAX = <BAX - <BAE = <CAX - <CAD = <DAX as required √ Or: Let bisector of A meet the circumcircle at X, √ Draw AP, BP; construct PY perpendicular to AB √ Triangle PBY is congruent to triangle PAY so <YPB = <YPA <YPA = <DCA Half of angle at centre = angle at circumf., same arc √ Triangles YPA and DCA are similar so <YAP = <DAC √ So <PAX = <BAX - <BAP = <CAX - <CAD = <DAX as required √ 9. Let P be any point on the parabola with equation x2 = 4cy and let F be the focus of the parabola at (0,c). Q is the point on the line segment FP which divides FP in the ratio 2:1 (Q is closer to P than to F). Find the equation of the locus traced by Q as P moves along the parabola. [4] Let Q be (x,y), P is (2a, a2/c) a point on the parabola. √ Q is 2/3 of the way between F and P, √ So x = (2/3)2a = 4a/3 and y = (2/3) (a2/c c) + c – = c/3 + 2 a2/3c √ Hence y = c/3 + (2/3c) (9x2/16) = 3x2/8c + c/3 √
8. ABC is an acute-angled triangle with circumcentre P. If AD is the altitude to BC, prove that the bisector of angle A bisects <DAP. [5] See attached diagrams 2. Either: Let bisector of Angle A meet the circumcircle at X, √ Let AP meet the circumcircle at E. Join BE. <ACD = <AEB since <s at circumf., same arc √ Line APE is a diameter so <ABE = 90 √ Triangles ABE and ADC have two pairs of equal angles, so are similar, Hence <BAE = <DAC √ So <EAX = <BAX - <BAE = <CAX - <CAD = <DAX as required √ Or: Let bisector of A meet the circumcircle at X, √ Draw AP, BP; construct PY perpendicular to AB √ Triangle PBY is congruent to triangle PAY so <YPB = <YPA <YPA = <DCA Half of angle at centre = angle at circumf., same arc √ Triangles YPA and DCA are similar so <YAP = <DAC √ So <PAX = <BAX - <BAP = <CAX - <CAD = <DAX as required √ 9. Let P be any point on the parabola with equation x2 = 4cy and let F be the focus of the parabola at (0,c). Q is the point on the line segment FP which divides FP in the ratio 2:1 (Q is closer to P than to F). Find the equation of the locus traced by Q as P moves along the parabola. [4] Let Q be (x,y), P is (2a, a2/c) a point on the parabola. √ Q is 2/3 of the way between F and P, √ So x = (2/3)2a = 4a/3 and y = (2/3) (a2/c – c) + c = c/3 + 2 a2/3c √ Hence y = c/3 + (2/3c) (9x2/16) = 3x2/8c + c/3 √
SENIOR_MATHS_COMP_2015_Prelim_Solns.docx