I have a hyperbola in the first quadrant opening to the right given with the equation $$x=\sqrt{19.8y^2-6.9y+0.7}$$ and I have to find the tangent line from the point $(0, 0.08)$ to the hyperbola (the tangent line with positive slope, so above the hyperbola if it’s opening to the right).
The problem is I don’t know what the tangent point is on the hyperbola because then I would just be able to find the slope and use $y=mx+b$ to find the equation of the tangent line.
My idea is since there are many lines that intersect the hyperbola from that point I have to maximize the slope between the point and the hyperbola and then I should be able to find the tangent line between $(0, 0.08)$ and the hyperbola.
But I don’t know how to find the maximum slope between the line that goes through the point and the hyperbola?
First consider implicit differentiation.
$$x^2= 19.8y^2 – 6.9y + 0.7$$
$$\to \frac{d}{dx} x^2 = \frac{d}{dx} (19.8y^2 – 6.9y + 0.7)$$
$$\to 2x = 2(19.8)y \frac{dy}{dx} – 6.9 \frac{dy}{dx}$$
$$\to 2x = (2(19.8)y – 6.9) \frac{dy}{dx}$$
$$\to \frac{2x}{2(19.8)y – 6.9} = \frac{dy}{dx}$$
There is some point on the graph call it $(x_1, y_1)$ s.t. the tangent line to the graph at that point passes through $(0,0.08)$.
Thus, we have the two equations:
$$x_1^2= 19.8y_1^2 – 6.9y_1 + 0.7$$
$$\frac{0.08 – y_1}{0 – x_1} = \frac{2x_1}{2(19.8)y_1 – 6.9}$$
Solve the equations* for $x_1$ and $y_1$ then find the equation of the line that passes through $(x_1, y_1)$ and $(0,0.08)$.
*You’ll probably get more than one solution. If so, do you know which one to pick?
I already gave answer to a similar question by @idknuttin which was closed as duplicate.
Below is the edited version of my original answer, which is, in essence, just an elaborate version of the answer by @BCLC.
I retype it here with minor edits for two main reasons:
$\hspace{2ex}$ a) OP of the second question seem to have troubles writing out explicit tangent line equation,
$\hspace{2ex}$ b) there was an algebra-relates mistake in my original answer
First, rewrite your equation:
\begin{align}
x&=\sqrt{19.8y^2-6.9y+0.7} & \iff&& x^2 & = 19.8y^2-6.9y+0.7
\end{align}
Second, take derivatives with respect to $x$ keeping in mind that $y = y(x)$:
\begin{align}
2x & = 39.6yy’ – 6.9y’ & \implies&& y’ &= \dfrac{2x}{39.6y-6.9}
\end{align}
Third, substitute expression for derivative $y'(x_0)$ to tangent line equation $%\dfrac{y – y_0}{x-x_0} = y'(x_0)$:
\begin{align}
\dfrac{y – y_0}{x-x_0} &= y'(x_0) = \dfrac{2x_0}{39.6y_0-6.9} %&\implies &&
%y – y_0 &= \dfrac{2x(x-x_0)}{39.6y-6.9} & \iff &&
%\bbox[1.5ex, border: solid 2pt #e10000]{(y – y_0)(39.6y-6.9) = 2x(x-x_0)}
\end{align}
where $(x_0,y_0)$ is the point of tangency.
Fourth, find point of tangency $(x_0,y_0)$ and value of derivative $y'(x_0)$ in three following steps:
tangent line passes through point $(x,y) = (0,0.08)$, thus we can substitute respective values into the tangent line equation:
\begin{align}
\dfrac{y_0 – 0.08}{x_0} &= \dfrac{2x_0}{39.6y_0-6.9} & \iff &&
2 x^2 &= 39.6 y^2-10.068 y+0.552
\end{align}
recall that the point of tangency $(x_0,y_0)$ also belongs to hyperbola, thus we can determine values of $x_0$ and $y_0$ by solving system of equations
\begin{align}
&\left\lbrace\begin{aligned}
x^2 &= 19.8y^2-6.9y+0.7 \\
2\hspace{0.125ex} x^2 &= 39.6 y^2-10.068 y+0.552
\end{aligned}\right.
&&\iff
\left\lbrace\begin{aligned}
x^2 – 19.8y^2 & = 0.7 – 6.9y \\
x^2 – 19.8y^2 & = 0.552 – 10.068 y
\end{aligned}\right.
\\[1.5ex] &&& \implies
6.9y – 0.7 = 5.034y – 0.276
\\[1.5ex] &&& \implies
\bbox[1.5ex,border:solid 2pt#e10000]{y_0 = 0.227224,\quad x_0 \approx 0.392993}
\end{align}
compute value of derivative at the point of tangency:
\begin{align}
y'(x_0) &= \left.\dfrac{2x_0}{39.6y_0-6.9}\;\right\rvert_{\scriptstyle {x_0 \,=\, 0.392993 \atop y_0 \,=\, 0.227224}} = 0.3746232729
%_{(0.392993,\,0.227224)} = 0.3746232729
\end{align}
Finally, write down explicitly the expression for the tangent line $y = y_0 + y'(x_0)(x-x_0)$:
\begin{align}
y = 0.227224 + 0.3746232729\hspace{0.25ex}(x – 0.392993)
\end{align}
Simplifying expression above, we get
\begin{align}
\bbox[2ex,border:solid 2pt #e10000]{y = 0.374623\hspace{0.125ex}x+0.0799997}
\end{align}
One can verify obtained results by plotting (e.g. in WolframAlpha) both hyperbola and tangent line: