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6561b 1) find the radius of the circle: 4x^2+4y^2=9 The equation becomes x2 + y2 = 9/4 = (3/2)2. \R = 3/2 = 1.5 Answer: A 2)what is the length of the major axis of the ellipse, 3x^2+4y^2=12 The equation becomes x2/4 + y2/3 = 1, or x2/22 + y2/(Ö3)2 = 1. The semi-major axis is a = 2. \ Major axis is 2a = 4. Answer: D 3)find the value of f(0)of the function: f(x)=3x^2+2x+5. f(0) = 0 + 0 + 5 = 5 Answer: B 4) Find the equation of a paraloba with its vertex at the point of origin and the focal point f(0,2). Such a parabola is described by x2 = 4ay - a normal parabola with the axis of symmetry along the y-axis, and its vertex at the origin. For x2 = 4ay, the focal point is given by f(0,a). \ a = 2. Answer: x2 = 8y 5) If the equation of the hyperbola is: 9x^2-64y^2=576 the point p(-8,0)is its The equation becomes x2/82 - y2/92 = 1, or x2/a2 - y2/b2 = 1. The vertices are given by x = ±a = ±8. So (-8,0) is the left-hand vertex. Answer: B 6) What are the polar coordinates of the point k(-1,- Ö3)? Polar coordinates are given by <r, q>. (x, y) = (-1,-Ö3), where x = rcosq, y = rsinq, and x2 + y2 = r2. (-1)2 + (-Ö3)2 = r2 1 + 3 = r2 4 = r2 r = 2 (r is always taken as +ve) tanq = (rsinq)/(rcosq) = (y/x) = -Ö3/-1 = Ö3. \ q = 60° = p/3. Answer: <r, q> = <2k, p/3> 7) Find the points of intersection of the straight line 7x+y=25 and the circle x^2+y^2=25 7x+y=25 gives y = 25 - 7x substitute for y = 25 - 7x into the equation for the circle. x2 + (25 - 7x)2 = 25 x2 + 625 - 350x + 49x2 = 25 50x2 - 350x + 600 = 0 x2 - 7x + 12 = 0 (x-3)(x-4) = 0 x = 3, or x = 4, and y = 25 - 7(3), or y = 25 - 7(4) x = 3,4 and y = 4, -3 The points of intersection are: (3,4) and (4,-3) 8) find the coordinates of the center of the circle x^2+y^2-4x+6y=12 Complete the square: x2 - 4x + y2 + 6y = 12 (x2 - 4x + 4) - 4 + (y2 + 6y + 9) - 9 = 12 (x-2)2 + (y + 3)2 = 12 + 4 + 9 = 25 (x-2)2 + (y + 3)2 = 52 Centre of circle is (h,k) = (2, -3) 9) Find the equation of the tangent to the ellipse x^2/9+y^2=1 at the point p(2, Ö5/3) Differentiating the ellipse gives: 2x/9 + 2y.y' = 0 y' = -x/(9y) At (x,y) = (2, Ö5/3), the slope, y' = -2/(9*Ö5/3) = -2/(3Ö5) Eqn of line is: (y-Ö5/3)/(x-2) = -2/(3Ö5) 3Ö5.y - 5 = -2x + 4 y = (9 - 2x)/ (3Ö5) 10) Which 1 of the following hyperbolas contains the point m(2Ö2,1)? ? 11) Find the coordinates of the focus of the conic section x^2+64y=0 This conic section is: x2 = 4ay, with a = -16 The focus is given by f = (0,a) = (0, -16) Answer: D 12)find the intersection points of the parabola y^2=-8x and the circle x^2+y^2-2x24=0 substitute for y2 = -8x into the equation for the circle. x2 - 8x - 2x - 24 = 0 x2 - 10x - 24 = 0 (x + 2)(x - 12) = 0 x = -2, or x = 12, x = -2 (x = -12 is a solution of the quadratic equation, but not of the problem, so is ignored - there is no point on the circle with an x-coordinate = 12) at x = -2, y2 = -8*(-2) = 16 y = ±4 The points of intersection are: (-2, 4), (-2, -4) 13) Find the equation of the circle which has its center at the point of origin & passes through the point p(2,1) The radius is given by r2 = x2 + y2 = 22 + 12 = 4 + 1 = 5 The equation of the circle is: x2 + y2 = 5 14)which of the following ellipses has its center at point c(2,1)? a)x^2+4y^2-4x-16y+4=0 b)4x^2+9y^2-16x-18y-11=0 c)9x^2+4y^2-36x-4y+4=0 d)16x^2+25y^2-36x+50y-164=0 Complete the square on b) 4x2+9y2-16x-18y-11=0 4x2 - 16x + 9y2 - 18y - 11 = 0 (4x2 - 16x + 16) - 16 + (9y2 - 18y + 9) - 9 - 11 = 0 (2x - 4)2 + (3y - 3)2 = 36 (x - 2)2/9 + (y - 1)2/4 = 1 It's centre is (h,k) = (2,1) Ans: b) 15)find the equations of asymptotes of the hyperbola:25x^2-16y^2=400 Rearranging the eqn: 16y2=25x2 - 400, or y2 = (5/4)2x2 - 25. As x gets very large, the -25 becomes less and less significant and the relation between x and y tends towards y2 = (5/4)2x2, or y = ±(5/4)x. The asymptotes are: y = -1.25x and y = 1.25x 16)what are the rectangular coordinates of the point(10,150degrees) x = rcosq = 10*cos(150) = 10*(-Ö3/2) = -5Ö3. Y = rsinq = 10*sin(150) = 10*( ½ ) = 5 The rectangular coords are (x,y) = (-5Ö3, 5) = (-0.866, 5) Ans: d) 17) find the slope of the tangent of the hyperbola x^2/16-y^2/4=1 at the point r(5,3/4) There's a mistake here. The point (5, -3/4) doesn't lie on the hyperbola. It should be the point (5, -3/2) differentiating the eqn for the hyperbola. x/8 - y.y'/2 = 0, y' = x/(4y) at (5, -3/2), y' = 5/(4*(-3/2)) = -5/6 Ans: b) 18)write in polar coordinates the equation y=3x+5 substitute for x = rcosq, and y = rsinq. rsinq = 3rcosq + 5 r(sinq - 3cosq) = 5 Ans: r = 5/( sinq - 3cosq) 19)find the zero point of the function f(x)=x-6 The function f is zero (has a zero point) when x = 6. Ans: c) 20)which one of the following straight lines is a tangent to the parabola y^2=20x hint: find which one of the following linear equations yields only one solution when solved simultaneously with the equation of the parabola. Substitute for y = -x - 5 into the eqn for the parabola. (-x-5)2 = 20x x2 + 10x + 25 = 20x x2 - 10x + 25 = 0 (x - 5)2 = 0 x = 5, only one (repeated) root Ans: d)