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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)