Topical Revision Worksheet 6 Coordinate Geometry Standard Graphs (LX)

Topical Revision Worksheet 6 Topic: Coordinate Geometry & Standard Graphs Question 1 [2008 EOY, Q9] marks The graph below shows part of a quadratic curve which represents the water level, y metres, of a river over a period of time in hours during the operation of a water dam nearby. At 2 pm and 8 pm, the water level was at the recommended level of 5 metres. Let x represents the time in hours since 12pm. (i) At what time did the water level reaches its minimum? Solution: ; at 5pm [1] (ii) Find the equation of the curve, expressing it in the form where a, h and k are real numbers. Solution: The vertex is at (5, 3) so Sub (2, 5) into the equation to get Hence [4] (iii) Hence, or otherwise, determine the height of the water level at 12 pm. Solution: Sub the height at 12 pm is metres [2] 5 Height of water, y (in metres) 4 3 8 6 7 Time, x (hours) 2 8

Question 2 [2008 EOY, Q10] marks Find the range of values of k if the line intersects the curve more than once. Solution: Equate the two equations to get For more than one root, Hence or [5] Question 3 [2008 EOY, Q12] marks The diagram below shows a circle with radius r units and its centre is . (i) Given that the circle passes through the origin, find the exact value of r. Solution: [2] (ii) The circle is reflected about the y-axis and then moved 5 units downwards. Write down the coordinates of the new centre of the circle. Solution: [1] (iii) Find the equation of the new circle in the form where g, f and c are real constants. Solution: [3] x y 0

Question 4 [2008 EOY, Q15] marks The figure below shows the graphs of and . The two curves meet at A and B. (i) Show that the coordinates of A are and find the coordinates of B. Solution: Solve to get or Sub to get hence A is (shown) and sub to get so B is [3] (ii) Find the length of AB. Solution: AB or 6.36 units [2] (iii) Find the equation of the perpendicular bisector of AB. Solution: Midpoint Gradient of perpendicular hence Perpendicular bisector of AB is [3] y x B A

Question 5 [2009 EOY, Q1] marks Find the coordinates of the points at which the straight line intersects the curve . Solution: Sub into : The coordinates of the points are (−1, −4) and (−2, −3). [4] Question 6 [2009 EOY, Q9] marks The equation of a circle A is . Find the coordinates of the centre of circle A and state its radius. Solution: A has centre (− 3, 2), radius = 3 [3]

Question 7 [2009 EOY, Q11] marks Solutions to this question by accurate drawing will not be accepted. In the figure, ABCD is a rhombus with vertices A (−4, 6) and C (4, −2). Given that the vertex B lies on the line , calculate the (i) equations of AC and BD, Solution: midpoint of AC = midpoint of BD = (0, 2) gradient of AC = Equation of AC: Gradient of BD = 1 Equation of BD: [3] (ii) coordinates of B and D, Solution: Coordinates of B: Solve and simultaneously. Solving, x = 2, y = 4 B = (2, 4) Use midpoint to calculate coordinates of D: [3] x y A (4, 6) B C (4, 2) D

(iii) equation of AB, Solution: Gradient of AB = Equation of AB: [2] (iv) area of the rhombus ABCD and hence find the perpendicular distance from B to the line CD. Solution: Area of ABCD = units 2. Let perpendicular distance from B to line CD be h. [4]

Question 8 [2010 EOY, Q4] marks Find the range of values of k such that the curve meets the line at two distinct points. Solution: Substitute y: For two distinct roots, Hence [5] Question 9 [2010 EOY, Q5] marks A quadratic curve is such that it is symmetrical about the line , where . Given that it cuts the x-axis when , and the maximum value of the curve is , (i) explain why the other x-intercept is , and Solution: The roots are equidistant from the line of symmetry, so the other x-intercept is at [1] (ii) find, in terms of x, y and k, the equation of the curve Solution: Vertex form: Sub to get [A1], hence OR Product form: Sub to get , hence [3] (cont’d) Given further that the distance between the two x-intercepts is 6 units, find the coordinates of the y-intercept. Solution: [B1] hence (using vertex form) Hence y-intercept [3]

Question 10 [2010 EOY, Q6] marks On separate axes, sketch the graphs of (i) for , and Solution: [G1] – Open downwards [G1] – Vertex in 2 nd quadrant [2] (ii) for , and Solution: [G1] – Open upwards [G1] – Vertex lies on y-axis [2] Question 11 [2010 EOY, Q7] marks The diagram below shows three small identical circles inscribed in a big circle centred at the origin. Each small circle has radius r units (where ) and one of the small circles has its centre at the point . The small circles are arranged with a rotational symmetry about the origin of order 3. (i) By considering the rotational symmetry of the diagram, explain fully why . Solution: Due to rotational symmetry of order 3, OA (the radius of the bigger circle) is part of the line of symmetry of the equilateral triangle ABC, and hence makes an angle of 30° [M1] with the x-axis (alternate angle with ). Hence, [2] O x y A B C

(ii) Write down the equation of the bigger circle, in terms of x, y and r. Solution: Radius of big circle units [B1] Hence or [2] Given further that , (iii) find the coordinates of A, the point of contact between the smaller circle and the bigger circle as shown in the diagram, and Solution: When , [B1] hence coordinates of A are [3] (iv) show that the area of the triangle ABC is units 2. Solution: Due to symmetry, units Hence area of triangle [3] Question 12 [2011 EOY, Q6] marks The equation of a circle is given by . (i) Find the centre and the radius of the circle. Solution: Centre , radius 5 . [4] (ii) Hence or otherwise, find the greatest possible value of y. Solution: By sketching the circle, or show : Greatest y is 1 [2]

Question 13 [2011 EOY, Q10a] marks A quadratic curve has a minimum point at and passes through . If the curve does not meet the line , find the range of values of m. Solution: Equation of the curve is At Since line does not meets the curve, [5] Question 14 [2011 EOY, Q14] marks Solutions to this question by accurate drawing will not be accepted. The diagram shows the quadrilateral ABCD. The coordinates of A and B are and respectively. Points C and D lie on the x-axis and (i) Find the equation of the perpendicular bisector of line AB. Solution: Midpoint of AB, Gradient of AB Equation of perpendicular of AB is: [3] (ii) Find the equation of AD and hence find the coordinates of D. Solution: Equation of AD is At D, , [2] (iii) If the area of is 14 square units, find the coordinates of C Solution: [1]