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This is homework, but the problem set is already due, but I don't understand how to get the answer to this problem and I'd really like to understand the thought process behind disseminating this problem.

    On your own paper, sketch the graph of the parabola 
y = x^2 +8
. On the same graph, plot the point 
(0, −3)
. Note that there are two tangent lines of y that pass through this point. (It may be useful to sketch them on your graph.)

Specifically, the tangent line of the parabola 
y = x^2 +8
 at the point 
(a, a^2+8)
 passes through the point 
(0, −3)
where a>0. The other tangent line passing through this point goes through 
(−a, a^2+8)

Use this information to find the number a.

Thanks everyone

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up vote 0 down vote accepted

As your exercise suggested, suppose that $(a,a^2+8)$ is a point of tangency. If $y=x^2+8$, then $\dfrac{dy}{dx}=2x$.

So the slope of the tangent line at $(2a,a^2+8)$ is $2a$.

It follows that the tangent line has equation $$y-(a^2+8)=2a(x-a).$$ We can rewrite this equation as $$y=2ax+8-a^2.\tag{$1$}$$ The above line passes through $(0,-3)$. Substitute these values of $x$ and $y$ in Equation $(1)$ . We get a quadratic equation in $a$. Solve for $a$. In this case, we get $-3=(2a)(0)+8-a^2$. So $a^2=11$ and therefore $a=\pm\sqrt{11}$. The $\pm$ makes sense, because of the symmetry of the picture about the $y$-axis: the points of tangency must be symmetrical about the $y$-axis.

For each of these values of $a$, substitute in Equation $(1)$ to find our two tangent lines.

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Thank you, this was helpful! – alvonellos Sep 13 '12 at 3:39

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