The curve C has equation $y = \frac{1}{3}x^2 + 8$ - Edexcel - A-Level Maths Pure - Question 11 - 2014 - Paper 2
Question 11
The curve C has equation $y = \frac{1}{3}x^2 + 8$.
The line L has equation $y = 3x + k$, where $k$ is a positive constant.
(a) Sketch C and L on separate diagrams,... show full transcript
Worked Solution & Example Answer:The curve C has equation $y = \frac{1}{3}x^2 + 8$ - Edexcel - A-Level Maths Pure - Question 11 - 2014 - Paper 2
Step 1
Sketch C and L on separate diagrams
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Answer
To sketch the curve C, we start by plotting the vertex at the point (0, 8) since the vertex form for the parabola is y=31x2+8.
Identify the intercepts for C:
The y-intercept occurs when x=0:
y=31(0)2+8=8, hence (0,8).
The x-intercepts are found by setting y=0:
Solve 0=31x2+8, which gives no real x-intercepts because 8 is positive.
Shape of C:
The graph is an upward-opening parabola, symmetric about the y-axis.
Sketch line L:
For the line y=3x+k, it will also cross the y-axis at the point (0,k).
The slope is positive, indicating the line will rise as x increases.
Identify points of intersection:
Line L will cut the y-axis at (0,k), and its x-intercept can be found by setting y=0:
0=3x+k⇒x=−3k, hence the intercept (−3k,0).
Step 2
find the value of k
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Answer
Since line L is tangent to curve C, at the point of tangency, both the y-values and the slopes of the two equations must be equal.
Equate curves:
Set the equations equal:
31x2+8=3x+k
Rearranging gives:
31x2−3x+(8−k)=0
Condition for tangency:
For tangency, the discriminant must be zero:
b2−4ac=0
Here, a=31,b=−3, and c=(8−k).
Calculate the discriminant:
(−3)2−4(31)(8−k)=0
Solving the equation gives:
9−34(8−k)=09=332−4k27=32−4k4k=32−27=5⇒k=45
Thus, the value of k is (1.25).