IB Mathematics: Analysis and Approaches Quiz: Conic Sections
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Conic SectionsQuestion 1 of 20

The horizontal line y=ky=k is tangent to the parabola y=x2−6x+13y = x^2 - 6x + 13. Find the value of kk.

44
33
−4-4
1313
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Conic Sections

Practice Conic Sections in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Conic Sections, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The horizontal line y=ky=k is tangent to the parabola y=x2−6x+13y = x^2 - 6x + 13. Find the value of kk.

  1. 44 (correct answer)
  2. 33
  3. −4-4
  4. 1313
Explanation: When you encounter a horizontal line tangent to a parabola, you're looking for the point where the parabola has a horizontal tangent line (where the derivative equals zero) and finding the y-coordinate at that point. For the parabola y=x2−6x+13y = x^2 - 6x + 13, first find where the tangent line is horizontal by taking the derivative: dydx=2x−6\frac{dy}{dx} = 2x - 6. Setting this equal to zero: 2x−6=02x - 6 = 0, so x=3x = 3. At x=3x = 3, the y-coordinate is: y=(3)2−6(3)+13=9−18+13=4y = (3)^2 - 6(3) + 13 = 9 - 18 + 13 = 4. Therefore, the horizontal tangent line is y=4y = 4, making k=4k = 4. You can verify this by completing the square: y=x2−6x+13=(x−3)2+4y = x^2 - 6x + 13 = (x-3)^2 + 4. This vertex form clearly shows the parabola's minimum point is at (3,4)(3, 4), confirming the horizontal tangent line is y=4y = 4. Looking at the incorrect answers: B) k=3k = 3 gives you the x-coordinate of the vertex instead of the y-coordinate—a common mix-up. C) k=−4k = -4 might result from a sign error when completing the square or substituting. D) k=13k = 13 is the y-intercept of the parabola (when x=0x = 0), but this isn't where the horizontal tangent occurs. Study tip: For parabola tangent problems, remember that horizontal tangent lines occur at the vertex. Use the derivative to find the x-coordinate, then substitute back to find the y-coordinate. Completing the square can provide a useful check of your work.

Question 2

A parabola has the equation y=−(x−3)2+4y = -(x-3)^2 + 4. A second parabola is formed by reflecting the first parabola in the x-axis. What is the equation of the second parabola?

  1. y=(x−3)2−4y = (x-3)^2 - 4 (correct answer)
  2. y=−(x+3)2+4y = -(x+3)^2 + 4
  3. y=−(x−3)2−4y = -(x-3)^2 - 4
  4. y=(x+3)2−4y = (x+3)^2 - 4
Explanation: When you encounter parabola transformations, focus on how each type of reflection affects the equation systematically. Reflecting across the x-axis changes the sign of the entire y-value, which means you multiply the right side of the equation by -1. Starting with y=−(x−3)2+4y = -(x-3)^2 + 4, reflecting this parabola in the x-axis transforms every point (x,y)(x,y) to (x,−y)(x,-y). This means the new equation becomes y=−[−(x−3)2+4]y = -[-(x-3)^2 + 4]. Distributing the negative sign: y=(x−3)2−4y = (x-3)^2 - 4. The parabola that opened downward with vertex at (3,4)(3,4) now opens upward with vertex at (3,−4)(3,-4). Looking at the incorrect options: Option B, y=−(x+3)2+4y = -(x+3)^2 + 4, represents a reflection across the y-axis (changing the sign inside the parentheses) rather than the x-axis. Option C, y=−(x−3)2−4y = -(x-3)^2 - 4, only changes the constant term's sign, which would be a vertical translation, not a reflection. Option D, y=(x+3)2−4y = (x+3)^2 - 4, combines both transformations incorrectly—it reflects across both axes when only x-axis reflection was requested. The correct answer is A: y=(x−3)2−4y = (x-3)^2 - 4. Strategy tip: Remember that x-axis reflections flip the sign of the entire right side of the equation, while y-axis reflections flip the sign of x inside parentheses. Always apply the transformation to the complete expression, not just individual terms.

Question 3

A parabola has the equation y=−(x−3)2+4y = -(x-3)^2 + 4. A second parabola is formed by reflecting the first parabola in the x-axis. What is the equation of the second parabola?

  1. y=(x−3)2−4y = (x-3)^2 - 4 (correct answer)
  2. y=−(x+3)2+4y = -(x+3)^2 + 4
  3. y=−(x−3)2−4y = -(x-3)^2 - 4
  4. y=(x+3)2−4y = (x+3)^2 - 4
Explanation: When you encounter parabola transformations, focus on how each type of reflection affects the equation systematically. Reflecting across the x-axis changes the sign of the entire y-value, which means you multiply the right side of the equation by -1. Starting with y=−(x−3)2+4y = -(x-3)^2 + 4, reflecting this parabola in the x-axis transforms every point (x,y)(x,y) to (x,−y)(x,-y). This means the new equation becomes y=−[−(x−3)2+4]y = -[-(x-3)^2 + 4]. Distributing the negative sign: y=(x−3)2−4y = (x-3)^2 - 4. The parabola that opened downward with vertex at (3,4)(3,4) now opens upward with vertex at (3,−4)(3,-4). Looking at the incorrect options: Option B, y=−(x+3)2+4y = -(x+3)^2 + 4, represents a reflection across the y-axis (changing the sign inside the parentheses) rather than the x-axis. Option C, y=−(x−3)2−4y = -(x-3)^2 - 4, only changes the constant term's sign, which would be a vertical translation, not a reflection. Option D, y=(x+3)2−4y = (x+3)^2 - 4, combines both transformations incorrectly—it reflects across both axes when only x-axis reflection was requested. The correct answer is A: y=(x−3)2−4y = (x-3)^2 - 4. Strategy tip: Remember that x-axis reflections flip the sign of the entire right side of the equation, while y-axis reflections flip the sign of x inside parentheses. Always apply the transformation to the complete expression, not just individual terms.

Question 4

The horizontal line y=ky=k is tangent to the parabola y=x2−6x+13y = x^2 - 6x + 13. Find the value of kk.

  1. 44 (correct answer)
  2. 33
  3. −4-4
  4. 1313
Explanation: When you encounter a horizontal line tangent to a parabola, you're looking for the point where the parabola has a horizontal tangent line (where the derivative equals zero) and finding the y-coordinate at that point. For the parabola y=x2−6x+13y = x^2 - 6x + 13, first find where the tangent line is horizontal by taking the derivative: dydx=2x−6\frac{dy}{dx} = 2x - 6. Setting this equal to zero: 2x−6=02x - 6 = 0, so x=3x = 3. At x=3x = 3, the y-coordinate is: y=(3)2−6(3)+13=9−18+13=4y = (3)^2 - 6(3) + 13 = 9 - 18 + 13 = 4. Therefore, the horizontal tangent line is y=4y = 4, making k=4k = 4. You can verify this by completing the square: y=x2−6x+13=(x−3)2+4y = x^2 - 6x + 13 = (x-3)^2 + 4. This vertex form clearly shows the parabola's minimum point is at (3,4)(3, 4), confirming the horizontal tangent line is y=4y = 4. Looking at the incorrect answers: B) k=3k = 3 gives you the x-coordinate of the vertex instead of the y-coordinate—a common mix-up. C) k=−4k = -4 might result from a sign error when completing the square or substituting. D) k=13k = 13 is the y-intercept of the parabola (when x=0x = 0), but this isn't where the horizontal tangent occurs. Study tip: For parabola tangent problems, remember that horizontal tangent lines occur at the vertex. Use the derivative to find the x-coordinate, then substitute back to find the y-coordinate. Completing the square can provide a useful check of your work.

Question 5

The line y=2x+cy = 2x + c is tangent to the circle x2+y2=20x^2 + y^2 = 20. Find the possible values of cc.

  1. c=±20c = \pm \sqrt{20}
  2. c=±5c = \pm 5
  3. c=±10c = \pm 10 (correct answer)
  4. c=±20c = \pm 20
Explanation: Substitute y=2x+cy = 2x + c into the circle's equation: x2+(2x+c)2=20x^2 + (2x+c)^2 = 20. Expand: x2+4x2+4cx+c2=20x^2 + 4x^2 + 4cx + c^2 = 20. Combine terms to form a quadratic in xx: 5x2+(4c)x+(c2−20)=05x^2 + (4c)x + (c^2 - 20) = 0. For the line to be tangent, this quadratic must have exactly one real solution, which means its discriminant Δ\Delta must be zero. Δ=B2−4AC=0\Delta = B^2 - 4AC = 0. Here A=5,B=4c,C=c2−20A=5, B=4c, C=c^2-20. So, (4c)2−4(5)(c2−20)=0(4c)^2 - 4(5)(c^2-20) = 0. This gives 16c2−20(c2−20)=016c^2 - 20(c^2-20) = 0, which simplifies to 16c2−20c2+400=016c^2 - 20c^2 + 400 = 0, or −4c2+400=0-4c^2 + 400 = 0. Thus 4c2=4004c^2 = 400, c2=100c^2 = 100, and c=±10c = \pm 10.

Question 6

A circle has its center at (2,−1)(2, -1) and is tangent to the line y=−4y = -4. Find the equation of the circle.

  1. (x−2)2+(y+1)2=3(x-2)^2 + (y+1)^2 = 3
  2. (x−2)2+(y+1)2=9(x-2)^2 + (y+1)^2 = 9 (correct answer)
  3. (x−2)2+(y+1)2=16(x-2)^2 + (y+1)^2 = 16
  4. (x+2)2+(y−1)2=9(x+2)^2 + (y-1)^2 = 9
Explanation: The center of the circle is (h,k)=(2,−1)(h,k) = (2, -1). The line y=−4y = -4 is a horizontal line. The radius rr of the circle is the perpendicular distance from the center to the tangent line. In this case, it is the vertical distance between y=−1y = -1 and y=−4y = -4. The distance is r=∣−1−(−4)∣=∣−1+4∣=3r = |-1 - (-4)| = |-1 + 4| = 3. The equation of the circle is (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2, which is (x−2)2+(y−(−1))2=32(x-2)^2 + (y-(-1))^2 = 3^2, or (x−2)2+(y+1)2=9(x-2)^2 + (y+1)^2 = 9.

Question 7

A parabola has x-intercepts at x=−1x=-1 and x=3x=3, and it passes through the point (1,−8)(1, -8). Which of the following is the equation of the parabola?

  1. y=x2−2x−3y = x^2 - 2x - 3
  2. y=2x2+4x−6y = 2x^2 + 4x - 6
  3. y=−2x2+4x+6y = -2x^2 + 4x + 6
  4. y=2x2−4x−6y = 2x^2 - 4x - 6 (correct answer)
Explanation: Since the x-intercepts are -1 and 3, the equation of the parabola can be written in factored form as y=a(x−(−1))(x−3)=a(x+1)(x−3)y = a(x - (-1))(x - 3) = a(x+1)(x-3). To find the value of aa, we use the given point (1,−8)(1, -8). Substitute x=1x=1 and y=−8y=-8 into the equation: −8=a(1+1)(1−3)-8 = a(1+1)(1-3). This simplifies to −8=a(2)(−2)-8 = a(2)(-2), so −8=−4a-8 = -4a, which gives a=2a=2. The equation is y=2(x+1)(x−3)y = 2(x+1)(x-3). Expanding this gives y=2(x2−3x+x−3)=2(x2−2x−3)=2x2−4x−6y = 2(x^2 - 3x + x - 3) = 2(x^2 - 2x - 3) = 2x^2 - 4x - 6.

Question 8

The points A(0, 5) and B(4, 1) lie on a circle. The center of the circle lies on the line x+y=4x+y=4. Find the coordinates of the center.

  1. (1,3)(1, 3)
  2. (1.5,2.5)(1.5, 2.5) (correct answer)
  3. (2.5,1.5)(2.5, 1.5)
  4. (3,1)(3, 1)
Explanation: The center of a circle lies on the perpendicular bisector of any of its chords. Let's find the equation of the perpendicular bisector of the chord AB. The midpoint of AB is M(0+42,5+12)=(2,3)(\frac{0+4}{2}, \frac{5+1}{2}) = (2, 3). The gradient of AB is mAB=1−54−0=−44=−1m_{AB} = \frac{1-5}{4-0} = \frac{-4}{4} = -1. The gradient of the perpendicular bisector is the negative reciprocal, m⊥=1m_{\perp} = 1. The equation of the perpendicular bisector is y−3=1(x−2)y-3 = 1(x-2), which simplifies to y=x+1y = x+1. The center lies on this line and also on the line x+y=4x+y=4. We solve the system of equations: substitute y=x+1y=x+1 into x+y=4x+y=4 to get x+(x+1)=4x + (x+1) = 4, which means 2x+1=42x+1=4, so 2x=32x=3 and x=1.5x=1.5. Then y=1.5+1=2.5y = 1.5 + 1 = 2.5. The center is at (1.5,2.5)(1.5, 2.5).

Question 9

What is the maximum value of the function f(x)=−2x2+8x−5f(x) = -2x^2 + 8x - 5?

  1. −5-5
  2. 22
  3. 33 (correct answer)
  4. 55
Explanation: The function f(x)=−2x2+8x−5f(x) = -2x^2 + 8x - 5 is a downward-opening parabola (since a=−2<0a = -2 < 0), so its maximum value occurs at the vertex. The x-coordinate of the vertex is given by xv=−b/(2a)x_v = -b/(2a). Here, a=−2a=-2 and b=8b=8, so xv=−8/(2⋅−2)=−8/−4=2x_v = -8/(2 \cdot -2) = -8/-4 = 2. The maximum value is the y-coordinate of the vertex, which is f(2)f(2). f(2)=−2(2)2+8(2)−5=−2(4)+16−5=−8+16−5=3f(2) = -2(2)^2 + 8(2) - 5 = -2(4) + 16 - 5 = -8 + 16 - 5 = 3. The maximum value is 3.

Question 10

Find the coordinates of the intersection points of the parabola y=x2−3x+5y = x^2 - 3x + 5 and the line y=x+2y = x + 2.

  1. (−1,1)(-1, 1) and (3,5)(3, 5)
  2. (1,3)(1, 3) only
  3. (1,3)(1, 3) and (3,5)(3, 5) (correct answer)
  4. (1,−3)(1, -3) and (3,−5)(3, -5)
Explanation: To find the intersection points, set the two equations equal to each other: x2−3x+5=x+2x^2 - 3x + 5 = x + 2. Rearrange the terms to form a quadratic equation: x2−4x+3=0x^2 - 4x + 3 = 0. Factor the quadratic: (x−1)(x−3)=0(x-1)(x-3) = 0. The solutions are x=1x=1 and x=3x=3. Substitute these x-values back into the line equation y=x+2y = x + 2 to find the corresponding y-coordinates. If x=1x=1, y=1+2=3y = 1+2 = 3. If x=3x=3, y=3+2=5y = 3+2 = 5. The intersection points are (1,3)(1, 3) and (3,5)(3, 5).

Question 11

What is the maximum value of the function f(x)=−2x2+8x−5f(x) = -2x^2 + 8x - 5?

  1. −5-5
  2. 22
  3. 33 (correct answer)
  4. 55
Explanation: The function f(x)=−2x2+8x−5f(x) = -2x^2 + 8x - 5 is a downward-opening parabola (since a=−2<0a = -2 < 0), so its maximum value occurs at the vertex. The x-coordinate of the vertex is given by xv=−b/(2a)x_v = -b/(2a). Here, a=−2a=-2 and b=8b=8, so xv=−8/(2⋅−2)=−8/−4=2x_v = -8/(2 \cdot -2) = -8/-4 = 2. The maximum value is the y-coordinate of the vertex, which is f(2)f(2). f(2)=−2(2)2+8(2)−5=−2(4)+16−5=−8+16−5=3f(2) = -2(2)^2 + 8(2) - 5 = -2(4) + 16 - 5 = -8 + 16 - 5 = 3. The maximum value is 3.

Question 12

The points A(0, 5) and B(4, 1) lie on a circle. The center of the circle lies on the line x+y=4x+y=4. Find the coordinates of the center.

  1. (1,3)(1, 3)
  2. (1.5,2.5)(1.5, 2.5) (correct answer)
  3. (2.5,1.5)(2.5, 1.5)
  4. (3,1)(3, 1)
Explanation: The center of a circle lies on the perpendicular bisector of any of its chords. Let's find the equation of the perpendicular bisector of the chord AB. The midpoint of AB is M(0+42,5+12)=(2,3)(\frac{0+4}{2}, \frac{5+1}{2}) = (2, 3). The gradient of AB is mAB=1−54−0=−44=−1m_{AB} = \frac{1-5}{4-0} = \frac{-4}{4} = -1. The gradient of the perpendicular bisector is the negative reciprocal, m⊥=1m_{\perp} = 1. The equation of the perpendicular bisector is y−3=1(x−2)y-3 = 1(x-2), which simplifies to y=x+1y = x+1. The center lies on this line and also on the line x+y=4x+y=4. We solve the system of equations: substitute y=x+1y=x+1 into x+y=4x+y=4 to get x+(x+1)=4x + (x+1) = 4, which means 2x+1=42x+1=4, so 2x=32x=3 and x=1.5x=1.5. Then y=1.5+1=2.5y = 1.5 + 1 = 2.5. The center is at (1.5,2.5)(1.5, 2.5).

Question 13

The vertex of the parabola y=x2−2kx+3y = x^2 - 2kx + 3 lies on the line y=x−ky = x - k. Find the positive value of kk.

  1. 11
  2. 3\sqrt{3} (correct answer)
  3. 5\sqrt{5}
  4. 33
Explanation: The x-coordinate of the vertex of a parabola y=ax2+bx+cy = ax^2+bx+c is given by xv=−b/2ax_v = -b/2a. For y=x2−2kx+3y = x^2 - 2kx + 3, a=1a=1 and b=−2kb=-2k, so xv=−(−2k)/(2⋅1)=kx_v = -(-2k)/(2 \cdot 1) = k. To find the y-coordinate of the vertex, substitute xv=kx_v=k into the parabola's equation: yv=k2−2k(k)+3=k2−2k2+3=−k2+3y_v = k^2 - 2k(k) + 3 = k^2 - 2k^2 + 3 = -k^2 + 3. The vertex is (k,−k2+3)(k, -k^2 + 3). Since the vertex lies on the line y=x−ky = x - k, its coordinates must satisfy the line's equation: −k2+3=k−k-k^2 + 3 = k - k. This simplifies to −k2+3=0-k^2 + 3 = 0, or k2=3k^2 = 3. The possible values for kk are k=±3k = \pm\sqrt{3}. The question asks for the positive value, which is k=3k = \sqrt{3}.

Question 14

A parabola has x-intercepts at x=−1x=-1 and x=3x=3, and it passes through the point (1,−8)(1, -8). Which of the following is the equation of the parabola?

  1. y=x2−2x−3y = x^2 - 2x - 3
  2. y=2x2+4x−6y = 2x^2 + 4x - 6
  3. y=−2x2+4x+6y = -2x^2 + 4x + 6
  4. y=2x2−4x−6y = 2x^2 - 4x - 6 (correct answer)
Explanation: Since the x-intercepts are -1 and 3, the equation of the parabola can be written in factored form as y=a(x−(−1))(x−3)=a(x+1)(x−3)y = a(x - (-1))(x - 3) = a(x+1)(x-3). To find the value of aa, we use the given point (1,−8)(1, -8). Substitute x=1x=1 and y=−8y=-8 into the equation: −8=a(1+1)(1−3)-8 = a(1+1)(1-3). This simplifies to −8=a(2)(−2)-8 = a(2)(-2), so −8=−4a-8 = -4a, which gives a=2a=2. The equation is y=2(x+1)(x−3)y = 2(x+1)(x-3). Expanding this gives y=2(x2−3x+x−3)=2(x2−2x−3)=2x2−4x−6y = 2(x^2 - 3x + x - 3) = 2(x^2 - 2x - 3) = 2x^2 - 4x - 6.

Question 15

Find the coordinates of the intersection points of the parabola y=x2−3x+5y = x^2 - 3x + 5 and the line y=x+2y = x + 2.

  1. (−1,1)(-1, 1) and (3,5)(3, 5)
  2. (1,3)(1, 3) only
  3. (1,3)(1, 3) and (3,5)(3, 5) (correct answer)
  4. (1,−3)(1, -3) and (3,−5)(3, -5)
Explanation: To find the intersection points, set the two equations equal to each other: x2−3x+5=x+2x^2 - 3x + 5 = x + 2. Rearrange the terms to form a quadratic equation: x2−4x+3=0x^2 - 4x + 3 = 0. Factor the quadratic: (x−1)(x−3)=0(x-1)(x-3) = 0. The solutions are x=1x=1 and x=3x=3. Substitute these x-values back into the line equation y=x+2y = x + 2 to find the corresponding y-coordinates. If x=1x=1, y=1+2=3y = 1+2 = 3. If x=3x=3, y=3+2=5y = 3+2 = 5. The intersection points are (1,3)(1, 3) and (3,5)(3, 5).

Question 16

The axis of symmetry of the parabola y=2x2+bx−7y = 2x^2 + bx - 7 is the line x=3x = 3. Find the value of bb.

  1. −12-12 (correct answer)
  2. −6-6
  3. 66
  4. 1212
Explanation: The equation for the axis of symmetry of a parabola y=ax2+bx+cy = ax^2 + bx + c is x=−b2ax = -\frac{b}{2a}. In this case, a=2a=2 and the axis of symmetry is x=3x=3. So we can set up the equation: 3=−b2(2)3 = -\frac{b}{2(2)}. This simplifies to 3=−b43 = -\frac{b}{4}. Multiplying both sides by -4 gives −12=b-12 = b.

Question 17

A circle has its center at (2,−1)(2, -1) and is tangent to the line y=−4y = -4. Find the equation of the circle.

  1. (x−2)2+(y+1)2=3(x-2)^2 + (y+1)^2 = 3
  2. (x−2)2+(y+1)2=9(x-2)^2 + (y+1)^2 = 9 (correct answer)
  3. (x−2)2+(y+1)2=16(x-2)^2 + (y+1)^2 = 16
  4. (x+2)2+(y−1)2=9(x+2)^2 + (y-1)^2 = 9
Explanation: The center of the circle is (h,k)=(2,−1)(h,k) = (2, -1). The line y=−4y = -4 is a horizontal line. The radius rr of the circle is the perpendicular distance from the center to the tangent line. In this case, it is the vertical distance between y=−1y = -1 and y=−4y = -4. The distance is r=∣−1−(−4)∣=∣−1+4∣=3r = |-1 - (-4)| = |-1 + 4| = 3. The equation of the circle is (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2, which is (x−2)2+(y−(−1))2=32(x-2)^2 + (y-(-1))^2 = 3^2, or (x−2)2+(y+1)2=9(x-2)^2 + (y+1)^2 = 9.

Question 18

The line y=2x+cy = 2x + c is tangent to the circle x2+y2=20x^2 + y^2 = 20. Find the possible values of cc.

  1. c=±20c = \pm \sqrt{20}
  2. c=±5c = \pm 5
  3. c=±10c = \pm 10 (correct answer)
  4. c=±20c = \pm 20
Explanation: Substitute y=2x+cy = 2x + c into the circle's equation: x2+(2x+c)2=20x^2 + (2x+c)^2 = 20. Expand: x2+4x2+4cx+c2=20x^2 + 4x^2 + 4cx + c^2 = 20. Combine terms to form a quadratic in xx: 5x2+(4c)x+(c2−20)=05x^2 + (4c)x + (c^2 - 20) = 0. For the line to be tangent, this quadratic must have exactly one real solution, which means its discriminant Δ\Delta must be zero. Δ=B2−4AC=0\Delta = B^2 - 4AC = 0. Here A=5,B=4c,C=c2−20A=5, B=4c, C=c^2-20. So, (4c)2−4(5)(c2−20)=0(4c)^2 - 4(5)(c^2-20) = 0. This gives 16c2−20(c2−20)=016c^2 - 20(c^2-20) = 0, which simplifies to 16c2−20c2+400=016c^2 - 20c^2 + 400 = 0, or −4c2+400=0-4c^2 + 400 = 0. Thus 4c2=4004c^2 = 400, c2=100c^2 = 100, and c=±10c = \pm 10.

Question 19

Find the shortest distance from the point P(8, 5) to the circle given by the equation (x−2)2+(y+3)2=4(x-2)^2 + (y+3)^2 = 4.

  1. 66
  2. 88 (correct answer)
  3. 1010
  4. 1212
Explanation: The circle has its center at C(2,−3)(2, -3) and a radius of r=4=2r = \sqrt{4} = 2. First, calculate the distance between the point P(8, 5) and the center of the circle C. Using the distance formula: d=(8−2)2+(5−(−3))2=62+82=36+64=100=10d = \sqrt{(8-2)^2 + (5 - (-3))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10. Since this distance (10) is greater than the radius (2), the point P is outside the circle. The shortest distance from an external point to a circle is the distance from the point to the center, minus the radius. Shortest distance = d−r=10−2=8d - r = 10 - 2 = 8.

Question 20

The vertex of the parabola y=x2−2kx+3y = x^2 - 2kx + 3 lies on the line y=x−ky = x - k. Find the positive value of kk.

  1. 11
  2. 3\sqrt{3} (correct answer)
  3. 5\sqrt{5}
  4. 33
Explanation: The x-coordinate of the vertex of a parabola y=ax2+bx+cy = ax^2+bx+c is given by xv=−b/2ax_v = -b/2a. For y=x2−2kx+3y = x^2 - 2kx + 3, a=1a=1 and b=−2kb=-2k, so xv=−(−2k)/(2⋅1)=kx_v = -(-2k)/(2 \cdot 1) = k. To find the y-coordinate of the vertex, substitute xv=kx_v=k into the parabola's equation: yv=k2−2k(k)+3=k2−2k2+3=−k2+3y_v = k^2 - 2k(k) + 3 = k^2 - 2k^2 + 3 = -k^2 + 3. The vertex is (k,−k2+3)(k, -k^2 + 3). Since the vertex lies on the line y=x−ky = x - k, its coordinates must satisfy the line's equation: −k2+3=k−k-k^2 + 3 = k - k. This simplifies to −k2+3=0-k^2 + 3 = 0, or k2=3k^2 = 3. The possible values for kk are k=±3k = \pm\sqrt{3}. The question asks for the positive value, which is k=3k = \sqrt{3}.