# How to find the two equations of the perpendiculars to a segment that are tangent to a hyperbola?

Equation of the segment : $2x + 4y-3 = 0$ Equation of the hyperbola : $7x^2 - 4y^2 =14$

How do you find the equation of the two linear functions that are both perpendicular to the segment and tangent to the hyperbola?

Thanks

The segment has an equation of $\,2x+4y-3=0\,$, hence the equation you are finding should be

$$4x-2y+c=0$$

where $\,c\,$ is a real constant

Now differentiate the hyperbola equation w.r.t. $\,x\,$

$$14x-8y\frac{dy}{dx}=0$$ $$\Rightarrow\quad\frac{dy}{dx}=\frac{7x}{4y}\qquad$$

Since the slope of $\,4x-2y+c=0\,$ is $\,2$, we have

$$\frac{dy}{dx}=\frac{7x}{4y}=2$$ $$\Rightarrow\quad\frac78x=y\qquad$$

Substitute $\,x\ \,\text{for}\,\ y\,$ in the hyperbola equation

$$7x^2-4\left(\frac78x\right)^2=14$$

$$\Rightarrow\quad x=\pm\frac43\sqrt2,\ \ \ y=\pm\frac76\sqrt2\quad$$

Use the values of $\,x\,$ and $\,y\,$ to calculate $\,c\,$, then we get

$$c=2y-4x=\pm3\sqrt2$$

Thus, the straight line that is perpendicular to the segment and tangent to the hyperbola has an equation of:

$$4x-2y+3\sqrt2=0$$ $$\text{or}\quad 4x-2y-3\sqrt2=0\qquad$$

• Your answer is a good one as it shows the algebraic methods through calculus and substitution to find the equations. In my answer above I didn't give the answer directly but more or less an easy method of how to find them by using the graphical method of first visually drawing it, then using vector notation to draw the vectors that satisfy the conditions through linear equations and their trigonometric relationships. I prefer to try and guide people in the right direction as opposed to just giving them the answer with the result. They need to do the work for good practice. I did up vote! Mar 5, 2017 at 4:31

Draw the graphs of both original equations for the segment and the hyperbola. Then visually draw the lines that satisfies being perpendicular to the segment and tangent to the hyperbola. Then from these visualizations you can use vector notation of the two lines you drew and from there once you have vectors you can generate points, perform dot products, trigonometric functions and linear approximation to generate the linear equations of these two lines in the form of $y = mx +b$

However you may need to use the equation $m = \frac{y_2 - y_1}{x_2 - x_1}$ to find its slope. But if you know that the slope of a line is understood as $\frac{rise}{run}$ you can use the unit circle coordinate pairs to know that any point on the unit pair has a point $(x,y)$ where they are defined as $(\cos\theta, \sin\theta)$ where $\theta$ is the angle above the $x-axis$ in standard form. Now when you have the right triangle that is generated by the $x-axis$ and the unit vector radius $\vec r = <1>$; its slope $\frac{rise}{run}$ is defined as $\frac{\delta y}{\delta x}$ which is the same as $\frac{\sin\theta}{\cos\theta} = \tan\theta$.

Simply put the tangent of the angle $\theta$ above the horizon or horizontal axis in the standard position or in the positive direction is the slope of the line of any linear equation. So when you look at a the linear equation $y = mx + b$ it can be substituted as $y = x*tan\theta + b$.

This all works between linear equations, trigonometric functions and the structure of the circle simply because the Pythagorean Theorem $A^2 + B^2 = C^2$ and the equation of a circle $X^2 + Y^2 = r^2$ are the same exact thing. The only difference between them is the variable notation used and the context in which they are used.

Pythagorean Theorem is normally associated and used with triangles, lines and vectors where the equation of circle is used to represent the dual function of the points on a circle according to the $x$ & $y$ coordinate.

The reason this relationship works is because a triangle consists of 3 vectors or lines with 3 points of intersection called vertices or singly called a vertex that generates an interior angle between to vectors. Due to this vector relationship you can take a single leg of a triangle that is a unit vector and rotate it around a single fixed point at either the head or the tail end of the vector and that full rotation of $360°$ or $2\pi$ radians gives you the unit circle.

Once find the points of intersection between your original graphs and the needed imposed lines. You can easily generate any linear equation or line with two points by using the form $y - y_1 = m(x - x_1)$ Once again you can substitute $m$ with $\tan\theta$.