how to maximize the slope in order to find the tangent line of hyperbola from a point?

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?

Solutions Collecting From Web of "how to maximize the slope in order to find the tangent line of hyperbola from a point?"

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:

  1. $$x_1^2= 19.8y_1^2 – 6.9y_1 + 0.7$$

  2. $$\frac{0.08 – y_1}{0 – x_1} = \frac{2x_1}{2(19.8)y_1 – 6.9}$$

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:

    1. 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}

    2. 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}

    3. 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:

enter image description here