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We start with the basic parabola y=x2y=x^{2}. a The points A(3,9)A(3,9) and B(−2,4)B(-2,4)...
Apr 13, 2024
We start with the basic parabola y=x2y=x^{2}. a The points A(3,9)A(3,9) and B(−2,4)B(-2,4) lie on the parabola y=x2y=x^{2}. Find the yy-axis intercept of the line ABA B. Find and prove a general result which includes this special case. b Again consider the points A(3,9)A(3,9) and B(−2,4)B(-2,4) on the parabola. The line OAO A joins the origin OO to the point AA. Draw the line through BB parallel to OAO A, and find the coordinates of the point where this line crosses the parabola again. Find and prove a general result which includes this special case. c Again consider A(3,9)A(3,9). Describe the relationship between the midpoints of all the line segments parallel to OAO A with both endpoints on the parabola. d Now consider C(−4,16)C(-4,16) and D(0,3)D(0,3). Find the coordinates of the point where the line CDC D crosses the parabola again. Try to generalise this example. e Investigate similar results for other parabolas such as y=12x2y=\frac{1}{2} x^{2}.
Solution by Steps
step 1
To find the y-axis intercept of the line AB, we need to find the equation of the line passing through points A(3,9) and B(-2,4)
step 2
The slope (m) of the line is calculated using the formula m=y2−y1x2−x1 m = \frac{y_2 - y_1}{x_2 - x_1}
step 3
Substituting the coordinates of A and B into the slope formula gives m=4−9−2−3=−5−5=1 m = \frac{4 - 9}{-2 - 3} = \frac{-5}{-5} = 1
step 4
With the slope and a point, we can use the point-slope form of the equation of a line: y−y1=m(x−x1) y - y_1 = m(x - x_1)
step 5
Substituting point A(3,9) and the slope into the point-slope form gives y−9=1(x−3) y - 9 = 1(x - 3)
step 6
Simplifying the equation gives y=x+6 y = x + 6
step 7
The y-axis intercept occurs when x=0 x = 0 , so substituting x=0 x = 0 into the equation gives y=0+6 y = 0 + 6
Answer
The y-axis intercept of the line AB is 6.
Key Concept
Finding the y-axis intercept of a line given two points
Explanation
The y-axis intercept is found by determining the equation of the line and then evaluating it at x=0 x = 0 .
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Solution by Steps
step 1
To find the coordinates of the intersection of the line parallel to OA through B with the parabola y=x2 y = x^2 , we first find the slope of OA
step 2
The slope of OA is the same as the slope of AB, which we previously found to be 1
step 3
The equation of the line through B(-2,4) with slope 1 is y−4=1(x+2) y - 4 = 1(x + 2)
step 4
Simplifying the equation gives y=x+6 y = x + 6 , which is the equation of the line parallel to OA through B
step 5
To find the intersection with the parabola, we set y=x2 y = x^2 equal to the equation of the line, x2=x+6 x^2 = x + 6
step 6
Rearranging the equation gives x2−x−6=0 x^2 - x - 6 = 0
step 7
Factoring the quadratic equation gives (x−3)(x+2)=0 (x - 3)(x + 2) = 0
step 8
Solving for x gives x=3 x = 3 or x=−2 x = -2 . Since B is at x=−2 x = -2 , the other intersection point is at x=3 x = 3
step 9
Substituting x=3 x = 3 into the equation of the line gives y=3+6=9 y = 3 + 6 = 9
Answer
The coordinates of the intersection of the line parallel to OA through B with the parabola again are (3,9).
Key Concept
Finding the intersection of a line and a parabola
Explanation
The intersection is found by setting the equation of the line equal to the equation of the parabola and solving for x, then finding the corresponding y value.
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Solution by Steps
step 1
To describe the relationship between the midpoints of all line segments parallel to OA with both endpoints on the parabola y=x2 y = x^2 , we consider the general form of such a line segment
step 2
Let the endpoints of the segment be P(x1,x12) P(x_1, x_1^2) and Q(x2,x22) Q(x_2, x_2^2) , where x1 x_1 and x2 x_2 are the x-coordinates of the points on the parabola
step 3
The midpoint M of the segment PQ has coordinates M(x1+x22,x12+x222) M\left(\frac{x_1 + x_2}{2}, \frac{x_1^2 + x_2^2}{2}\right)
step 4
Since the line segments are parallel to OA, the slope of PQ is the same as the slope of OA, which is 1
step 5
The slope of PQ can also be expressed as x22−x12x2−x1 \frac{x_2^2 - x_1^2}{x_2 - x_1} , which simplifies to x2+x1 x_2 + x_1 due to the difference of squares
step 6
Since the slope is 1, we have x2+x1=1⋅(x2−x1) x_2 + x_1 = 1 \cdot (x_2 - x_1) , which simplifies to x2+x1=x2−x1 x_2 + x_1 = x_2 - x_1
step 7
Solving for x1 x_1 gives x1=0 x_1 = 0 , which implies that the midpoint of any such segment has an x-coordinate of 0
Answer
The midpoints of all line segments parallel to OA with both endpoints on the parabola y=x2 y = x^2 have an x-coordinate of 0.
Key Concept
Relationship between midpoints of line segments on a parabola
Explanation
The midpoints of all such line segments have an x-coordinate of 0 because the segments are parallel to a line with slope 1, and the parabola is symmetric about the y-axis.
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Solution by Steps
step 1
To find the coordinates of the point where the line CD crosses the parabola y=x2 y = x^2 again, we first find the equation of the line CD
step 2
The slope (m) of line CD is calculated using the formula m=y2−y1x2−x1 m = \frac{y_2 - y_1}{x_2 - x_1}
step 3
Substituting the coordinates of C(-4,16) and D(0,3) into the slope formula gives m=3−160−(−4)=−134 m = \frac{3 - 16}{0 - (-4)} = \frac{-13}{4}
step 4
With the slope and a point, we can use the point-slope form of the equation of a line: y−y1=m(x−x1) y - y_1 = m(x - x_1)
step 5
Substituting point D(0,3) and the slope into the point-slope form gives y−3=−134(x−0) y - 3 = \frac{-13}{4}(x - 0)
step 6
Simplifying the equation gives y=−134x+3 y = \frac{-13}{4}x + 3
step 7
To find the intersection with the parabola, we set y=x2 y = x^2 equal to the equation of the line, x2=−134x+3 x^2 = \frac{-13}{4}x + 3
step 8
Rearranging the equation gives x2+134x−3=0 x^2 + \frac{13}{4}x - 3 = 0
step 9
Solving this quadratic equation will give us the x-coordinates of the intersection points
step 10
Since C is at x=−4 x = -4 , we need to find the other solution for x
step 11
Using the quadratic formula, x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , we can find the other x-coordinate
step 12
Substituting a=1 a = 1 , b=134 b = \frac{13}{4} , and c=−3 c = -3 into the quadratic formula and solving for x gives us the other x-coordinate
step 13
Once we have the x-coordinate, we substitute it back into the equation of the line to find the corresponding y-coordinate
Answer
The coordinates of the point where the line CD crosses the parabola again can be found by solving the quadratic equation and substituting the x-coordinate into the line's equation.
Key Concept
Finding the intersection of a line and a parabola
Explanation
The intersection is found by setting the equation of the line equal to the equation of the parabola and solving for x, then finding the corresponding y value.
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Solution by Steps
step 1
To investigate the line intercepts and relationships for the parabola y=12x2 y = \frac{1}{2} x^2 , we consider the general form of a line y = mx + b and its intersections with the parabola
step 2
Setting the equation of the line equal to the parabola gives 12x2=mx+b \frac{1}{2} x^2 = mx + b
step 3
Rearranging the equation gives 12x2−mx−b=0 \frac{1}{2} x^2 - mx - b = 0
step 4
Solving this quadratic equation will give us the x-coordinates of the intersection points
step 5
Using the quadratic formula, x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , we can find the x-coordinates
step 6
Substituting a=12 a = \frac{1}{2} , b=−m b = -m , and c=−b c = -b into the quadratic formula and solving for x gives us the x-coordinates of the intercepts
step 7
Once we have the x-coordinates, we substitute them back into the equation of the line to find the corresponding y-coordinates
Answer
The line intercepts and relationships for the parabola y=12x2 y = \frac{1}{2} x^2 can be found by solving the quadratic equation and substituting the x-coordinates into the line's equation.
Key Concept
Investigating line intercepts with a parabola
Explanation
The intercepts are found by setting the equation of the line equal to the equation of the parabola and solving for x, then finding the corresponding y values.
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