Question

The solutions (x,y) of the equation x² + 16y² = 16 form an ellipse as pictured below.Consider the point P as pictured, with x-coordinate 3.

(a) Let h be a small non-zero number and form the point Q with x-coordinate 3+h, as pictured. The slope of the secant line through PQ, denoted s(h), is given by the formula =         ?

(b) Rationalize the numerator of your formula in (a) to rewrite the expression so that it looks like f(h)/g(h), subject to these two conditions: (1) thenumerator f(h) defines a line of slope -1, (2) the function f(h)/g(h) is defined for h=0. When you do this

f(h)=         

g(h)=         

(c) The slope of the tangent line to the ellipse at the point P is

=         

Answer 1

Consider the equation,

$$ \begin{aligned} x^{2}+16 y^{2} &=16 \\ y^{2} &=\frac{16-x^{2}}{16} \end{aligned} $$

\(y=\frac{\sqrt{16-x^{2}}}{4}\) (Consider the region in the first quadrant) Let \(y=F(x)=\frac{\sqrt{16-x^{2}}}{4}\)

The \(x\) -coordinate of the point \(P\) is 3 and the \(x\) -coordinate of \(Q\) is \(3+h\)

Therefore, the points are

$$ P(3, F(3)) \text { and } Q(3+h, F(3+h)) $$

(a) The slope of the secant line \(P Q\) is

$$ \begin{aligned} s(h) &=\frac{F(3+h)-F(3)}{3+h-3} \\ &=\frac{\sqrt{16-(3+h)^{2}}}{4}-\frac{\sqrt{16-3^{2}}}{4} \\ &=\frac{\sqrt{7-6 h-h^{2}}-\sqrt{7}}{4 h} \end{aligned} $$

(b) Rationalize the num erator of \(s(h)\)

$$ \begin{aligned} s(h) &=\frac{\sqrt{7-6 h-h^{2}}-\sqrt{7}}{4 h} \times \frac{\sqrt{7-6 h-h^{2}}+\sqrt{7}}{\sqrt{7-6 h-h^{2}}+\sqrt{7}} \\ &=\frac{\left(7-6 h-h^{2}\right)-(7)}{4 h\left(\sqrt{7-6 h-h^{2}}+\sqrt{7}\right)} \\ &=\frac{-6 h-h^{2}}{4 h\left(\sqrt{7-6 h-h^{2}}+\sqrt{7}\right)} \\ &=\frac{-6-h}{4\left(\sqrt{7-6 h-h^{2}}+\sqrt{7}\right)} \end{aligned} $$

Now rewrite the expression into: (1) the numerator \(f(h)\) defines a line of slope -1 and; (2) the function \(\frac{f(h)}{g(h)}\) is defined for \(h=0\). Clearly the above function \(s(h)\) satisfies the conditions. So, \(\quad f(h)=-6-h\) and

$$ g(h)=\sqrt{4\left(\sqrt{7-6 h-h^{2}}+\sqrt{7}\right)} $$

\((c)\)

The slope of the tangent line to the ellipse at \(P\) is

$$ \begin{aligned} \lim _{h \rightarrow 0} s(h) &=\lim _{h \rightarrow 0} \frac{-6-h}{4\left(\sqrt{7-6 h-h^{2}}+\sqrt{7}\right)} \\ &=\frac{-6-0}{4\left(\sqrt{7-6(0)-(0)^{2}}+\sqrt{7}\right)} \\ &=\frac{-6}{4(\sqrt{7}+\sqrt{7})} \\ &=\frac{-6}{4(2 \sqrt{7})} \\ &=\frac{-3}{4 \sqrt{7}} \end{aligned} $$