I cannot tell whether you are taking calculus or are still
in algebra. So I'll do it both ways.
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If you are taking algebra, this is the way to approach it:
Here is the graph of y² = 5x + 10
Here is a graph of the line y = x + 2.5 where k=2.5
As you can see that line is not quite tangent to the parabola,
because it intersects the parabola in two places. So k=2.5 is
not quite large enough.
Here is a graph of the line y = x + 4 where k = 4
And as you see that graph doesn't intersect the parabola at all.
So k=4 is too large.
We want to find the exact value of k, so that the line whose equation
is y = x + k only has one point of intersection with the graph. So we
solve the system of equations:
y² = 5x + 10
y = x + k
and find k so that there is just 1 solution, not 2 or 1.
So we substitute (x + k) for x in the first equation:
(x + k)² = 5x + 10
x² + 2kx + k² = 5x + 10
x² + 2kx - 5x + k² - 10 = 0
x² + (2k-5)x + (k²-10) = 0
That will have one solution when the discriminant b²-4ac = 0
b²-4ac = (2k-5)² - 4(1)(k²-10)
= 4k²-20k+25-4(k²-10)
= 4k²-20k+25-4k²+ 40
= -20k+65
So we set that = 0
20k+65= 0
2k=-65
k=
k=
That's the answer. Now as a graphical check,
when we graph y = x -
we get:
We see that the line y = x -
is exactly tangent to
the graph of y² = 5x + 10
So we are satisfied with our answer k =
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Now if you are taking calculus, you will find the derivatives of
both the line and the parabola and find where they are equal:
The derivative of
y = x + k is
The derivative of
y² = 5x + 10
2y
= 5
=
We set the derivatives equal
That is the y-coordinate of the point of tangency.
Now we substitute that in
y² = 5x + 10
(
)² = 5x + 10
= 5x + 10
25 = 20x + 40
-15 = 20x
= x
= x
That is the x-coordinate of the point of tangency.
So the point of tangency is (
,
)
Substituting that in the equation of the line:
y = x + k
= (((-3/4}}} + k
= k
= k
= k
Either way we get k =
Edwin