(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Draw a diagram to show that there are two lines tangent to both of the parabolas

[itex]y = - x^{2}[/itex] (1)

and

[itex]y = 4 + x^{2}[/itex] (2)

Find the coordinates of the four points at which these tangents touch the parabolas.

2. Relevant equations

[itex]y - y_{o} = m(x - x_{o})[/itex]

3. The attempt at a solution

In creating a diagram, it appears that for such a scenario to occur one of the two lines must be tangent to one side (right will be used) of the parabola opening up (2) and then tangent to the other side (left), but of the parabola opening down (1) -- let this be line 1. Similarly, for the other line, it must be tangent to the other side (left) of (2), and then tangent to the other side (right) of (1) -- let this be line 2. So, it might appear that there are only two x-values for all four points (one negative, one positive) and two y-values (one negative, one positive), being that both curves are symmetrical at x = 0.

In theory, I believe that for a line to be tangent to both parabolas at two points x = a [on (2)] and x = b [on (1)] (that is, considering only line 1), it must have only one slope. Therefore, letting (1) equal to f(x), and (2) equal to g(x),

[itex]D_{x}g(a) = D_{x}f(b)[/itex]

[itex]2a = -2b[/itex]

[itex]a = -b[/itex]

So, this scenario is true when the x-coordinates are symmetrical about x = 0.

On line 1 tangent to (2) in the first quadrant at P(a, c) and tangent to (1) in the third quadrant at Q(b, d), the points can be expressed differently.

[itex]P(a, c) = P(a, 4 + a^{2})[/itex]

[itex]Q(b, d) = Q(-a, -[-a]^{2}) = Q(-a, -a^{2})[/itex] -- [itex]b = - a[/itex]

On line 2 tangent to (2) in the second quadrant at U(e, f) and tangent to (1) in the fourth quadrant at S(g, h), the points can be express differently.

[itex]U(e, f) = U(-a, 4 + [-a]^{2}) = U(-a, 4 + a^{2})[/itex]

[itex]S(g, h) = S(a, -a^{2})[/itex]

So, the slope of line PQ should be equal to either of the derivatives (the derivative of either curve) and the slope of line US should also be equal to either of the derivatives.

For line one,

[itex]\frac{4 + a^{2} + a^{2}}{a + a} = 2a[/itex]

[itex]\frac{4 + 2a^{2}}{2a} = 2a[/itex]

[itex]4a^{2} = 4 + 2a^{2}[/itex]

[itex]2a^{2} = 2 + a^{2}[/itex]

[itex]0 = 2 - a^{2}[/itex]

[itex]a = \pm \sqrt{2}[/itex]

It appears the plus or minus further confirms the symmetry.

For line two,

[itex]\frac{4 + a^{2} + a^{2}}{-a - a} = 2a[/itex]

[itex]\frac{4 + 2a^{2}}{-2a} = 2a[/itex]

[itex]-4a^{2} = 4 + 2a^{2}[/itex]

[itex]-2a^{2} = 2 + a^{2}[/itex]

[itex]0 = 2 + 3a^{2}[/itex]

[itex]a = \pm \sqrt{-\frac{2}{3}}[/itex]

Which is not defined (I found this odd).

Therefore, the four points are,

[itex]P(\sqrt{2}, 4 + [\sqrt{2}]^{2}) = P(\sqrt{2}, 6)[/itex] -- Because P is in the first quadrant, touching curve (2)

[itex]Q(-\sqrt{2}, -[-\sqrt{2}]^{2}) = Q(-\sqrt{2}, -2)[/itex] -- Because Q is in the third quadrant, touching curve (1)

[itex]U(-\sqrt{2}, 4 + [-\sqrt{2}]^{2}) = U(-\sqrt{2}, 6)[/itex] -- Because U is in the second quadrant, touching curve (2)

[itex]S(\sqrt{2}, -[\sqrt{2}]^{2}) = S(\sqrt{2}, -2)[/itex] -- Because S is in the fourth quadrant, touching curve (1)

And, the points illustrate symmetry about x = 0 (spefically, P and U, and, Q and S).

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Much appreciation for any help!

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# Homework Help: Find the four points at which two lines are tangent to both parabolas

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