Conic Sections - IB Mathematics Analysis and Approaches

The line $y = 2x + c$ is tangent to the circle $x^2 + y^2 = 20$ . Find the possible values of $c$ .
Answer: $c = \pm 10$ Substitute $y = 2x + c$ into the circle's equation: $x^2 + (2x+c)^2 = 20$ . Expand: $x^2 + 4x^2 + 4cx + c^2 = 20$ . Combine terms to form a quadratic in $x$ : $5x^2 + (4c)x + (c^2 - 20) = 0$ . For the line to be tangent, this quadratic must have exactly one real solution, which means its discriminant $\Delta$ must be zero. $\Delta = B^2 - 4AC = 0$ . Here $A=5, B=4c, C=c^2-20$ . So, $(4c)^2 - 4(5)(c^2-20) = 0$ . This gives $16c^2 - 20(c^2-20) = 0$ , which simplifies to $16c^2 - 20c^2 + 400 = 0$ , or $-4c^2 + 400 = 0$ . Thus $4c^2 = 400$ , $c^2 = 100$ , and $c = \pm 10$ .
A parabola has the equation $y = -(x-3)^2 + 4$ . A second parabola is formed by reflecting the first parabola in the x-axis. What is the equation of the second parabola?
Answer: $y = (x-3)^2 - 4$ 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 $y = -(x-3)^2 + 4$ , reflecting this parabola in the x-axis transforms every point $(x,y)$ to $(x,-y)$ . This means the new equation becomes $y = -[-(x-3)^2 + 4]$ . Distributing the negative sign: $y = (x-3)^2 - 4$ . The parabola that opened downward with vertex at $(3,4)$ now opens upward with vertex at $(3,-4)$ . Looking at the incorrect options: Option B, $y = -(x+3)^2 + 4$ , represents a reflection across the y-axis (changing the sign inside the parentheses) rather than the x-axis. Option C, $y = -(x-3)^2 - 4$ , only changes the constant term's sign, which would be a vertical translation, not a reflection. Option D, $y = (x+3)^2 - 4$ , combines both transformations incorrectly—it reflects across both axes when only x-axis reflection was requested. The correct answer is A: $y = (x-3)^2 - 4$ . 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.
Find the coordinates of the intersection points of the parabola $y = x^2 - 3x + 5$ and the line $y = x + 2$ .
Answer: $(1, 3)$ and $(3, 5)$ To find the intersection points, set the two equations equal to each other: $x^2 - 3x + 5 = x + 2$ . Rearrange the terms to form a quadratic equation: $x^2 - 4x + 3 = 0$ . Factor the quadratic: $(x-1)(x-3) = 0$ . The solutions are $x=1$ and $x=3$ . Substitute these x-values back into the line equation $y = x + 2$ to find the corresponding y-coordinates. If $x=1$ , $y = 1+2 = 3$ . If $x=3$ , $y = 3+2 = 5$ . The intersection points are $(1, 3)$ and $(3, 5)$ .
Find the shortest distance from the point P(8, 5) to the circle given by the equation $(x-2)^2 + (y+3)^2 = 4$ .
Answer: $8$ The circle has its center at C $(2, -3)$ and a radius of $r = \sqrt{4} = 2$ . First, calculate the distance between the point P(8, 5) and the center of the circle C. Using the distance formula: $d = \sqrt{(8-2)^2 + (5 - (-3))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ . 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 = $d - r = 10 - 2 = 8$ .
The vertex of the parabola $y = x^2 - 2kx + 3$ lies on the line $y = x - k$ . Find the positive value of $k$ .
Answer: $\sqrt{3}$ The x-coordinate of the vertex of a parabola $y = ax^2+bx+c$ is given by $x_v = -b/2a$ . For $y = x^2 - 2kx + 3$ , $a=1$ and $b=-2k$ , so $x_v = -(-2k)/(2 \cdot 1) = k$ . To find the y-coordinate of the vertex, substitute $x_v=k$ into the parabola's equation: $y_v = k^2 - 2k(k) + 3 = k^2 - 2k^2 + 3 = -k^2 + 3$ . The vertex is $(k, -k^2 + 3)$ . Since the vertex lies on the line $y = x - k$ , its coordinates must satisfy the line's equation: $-k^2 + 3 = k - k$ . This simplifies to $-k^2 + 3 = 0$ , or $k^2 = 3$ . The possible values for $k$ are $k = \pm\sqrt{3}$ . The question asks for the positive value, which is $k = \sqrt{3}$ .
A parabola has x-intercepts at $x=-1$ and $x=3$ , and it passes through the point $(1, -8)$ . Which of the following is the equation of the parabola?
Answer: $y = 2x^2 - 4x - 6$ Since the x-intercepts are -1 and 3, the equation of the parabola can be written in factored form as $y = a(x - (-1))(x - 3) = a(x+1)(x-3)$ . To find the value of $a$ , we use the given point $(1, -8)$ . Substitute $x=1$ and $y=-8$ into the equation: $-8 = a(1+1)(1-3)$ . This simplifies to $-8 = a(2)(-2)$ , so $-8 = -4a$ , which gives $a=2$ . The equation is $y = 2(x+1)(x-3)$ . Expanding this gives $y = 2(x^2 - 3x + x - 3) = 2(x^2 - 2x - 3) = 2x^2 - 4x - 6$ .
What is the maximum value of the function $f(x) = -2x^2 + 8x - 5$ ?
Answer: $3$ The function $f(x) = -2x^2 + 8x - 5$ is a downward-opening parabola (since $a = -2 < 0$ ), so its maximum value occurs at the vertex. The x-coordinate of the vertex is given by $x_v = -b/(2a)$ . Here, $a=-2$ and $b=8$ , so $x_v = -8/(2 \cdot -2) = -8/-4 = 2$ . The maximum value is the y-coordinate of the vertex, which is $f(2)$ . $f(2) = -2(2)^2 + 8(2) - 5 = -2(4) + 16 - 5 = -8 + 16 - 5 = 3$ . The maximum value is 3.
The horizontal line $y=k$ is tangent to the parabola $y = x^2 - 6x + 13$ . Find the value of $k$ .
Answer: $4$ 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 $y = x^2 - 6x + 13$ , first find where the tangent line is horizontal by taking the derivative: $\frac{dy}{dx} = 2x - 6$ . Setting this equal to zero: $2x - 6 = 0$ , so $x = 3$ . At $x = 3$ , the y-coordinate is: $y = (3)^2 - 6(3) + 13 = 9 - 18 + 13 = 4$ . Therefore, the horizontal tangent line is $y = 4$ , making $k = 4$ . You can verify this by completing the square: $y = x^2 - 6x + 13 = (x-3)^2 + 4$ . This vertex form clearly shows the parabola's minimum point is at $(3, 4)$ , confirming the horizontal tangent line is $y = 4$ . Looking at the incorrect answers: B) $k = 3$ gives you the x-coordinate of the vertex instead of the y-coordinate—a common mix-up. C) $k = -4$ might result from a sign error when completing the square or substituting. D) $k = 13$ is the y-intercept of the parabola (when $x = 0$ ), 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.
The axis of symmetry of the parabola $y = 2x^2 + bx - 7$ is the line $x = 3$ . Find the value of $b$ .
Answer: $-12$ The equation for the axis of symmetry of a parabola $y = ax^2 + bx + c$ is $x = -\frac{b}{2a}$ . In this case, $a=2$ and the axis of symmetry is $x=3$ . So we can set up the equation: $3 = -\frac{b}{2(2)}$ . This simplifies to $3 = -\frac{b}{4}$ . Multiplying both sides by -4 gives $-12 = b$ .
The points A(0, 5) and B(4, 1) lie on a circle. The center of the circle lies on the line $x+y=4$ . Find the coordinates of the center.
Answer: $(1.5, 2.5)$ 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 $(\frac{0+4}{2}, \frac{5+1}{2}) = (2, 3)$ . The gradient of AB is $m_{AB} = \frac{1-5}{4-0} = \frac{-4}{4} = -1$ . The gradient of the perpendicular bisector is the negative reciprocal, $m_{\perp} = 1$ . The equation of the perpendicular bisector is $y-3 = 1(x-2)$ , which simplifies to $y = x+1$ . The center lies on this line and also on the line $x+y=4$ . We solve the system of equations: substitute $y=x+1$ into $x+y=4$ to get $x + (x+1) = 4$ , which means $2x+1=4$ , so $2x=3$ and $x=1.5$ . Then $y = 1.5 + 1 = 2.5$ . The center is at $(1.5, 2.5)$ .
A circle has its center at $(2, -1)$ and is tangent to the line $y = -4$ . Find the equation of the circle.
Answer: $(x-2)^2 + (y+1)^2 = 9$ The center of the circle is $(h,k) = (2, -1)$ . The line $y = -4$ is a horizontal line. The radius $r$ of the circle is the perpendicular distance from the center to the tangent line. In this case, it is the vertical distance between $y = -1$ and $y = -4$ . The distance is $r = |-1 - (-4)| = |-1 + 4| = 3$ . The equation of the circle is $(x-h)^2 + (y-k)^2 = r^2$ , which is $(x-2)^2 + (y-(-1))^2 = 3^2$ , or $(x-2)^2 + (y+1)^2 = 9$ .

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