# Question #a51c8

Oct 10, 2017

The point $\left(1 , 4\right)$ has a horizontal tangent line.
The point $\left(0 , 5\right)$ has a vertical tangent line.

#### Explanation:

Firstly, here are the steps for the differentiation.

Differentiate everything (implicitly) with respect to $x$.

$2 x - 2 \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 6 - 10 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Bring all the terms containing $\frac{\mathrm{dy}}{\mathrm{dx}}$ to one side of the equation.

$2 x - 2 y + 6 = 2 x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 10 \frac{\mathrm{dy}}{\mathrm{dx}}$

Remove the common multiple of 2.

$x - y + 3 = x \frac{\mathrm{dy}}{\mathrm{dx}} - y \frac{\mathrm{dy}}{\mathrm{dx}} + 5 \frac{\mathrm{dy}}{\mathrm{dx}}$

Make $\frac{\mathrm{dy}}{\mathrm{dx}}$ the subject of formula.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x - y + 3}{x - y + 5}$

Which is what you have gotten.

To find the point(s) with horizontal tangent line, you should look for points $\left(x , y\right)$ such that $\frac{\mathrm{dy}}{\mathrm{dx}}$ is zero.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x - y + 3}{x - y + 5} = 0$

which simplifies to

$y = x + 3$

Bear in mind that the point $\left(x , y\right)$ also has to satisfy the condition of lying on the parabola.

${x}^{2} - 2 x y + {y}^{2} + 6 x - 10 y + 25 = 0$

Solve this simultaneous equation by substitution.

${x}^{2} - 2 x \left(x + 3\right) + {\left(x + 3\right)}^{2} + 6 x - 10 \left(x + 3\right) + 25 = 0$

Expanding the terms reveal that the $x$ squared terms cancel out.

$4 - 4 x = 0$

$x = 1$

$y = 1 + 3 = 4$

The point $\left(1 , 4\right)$ has a horizontal tangent line. The equation of the tangent line is $y = 4$.

To find the point(s) with vertical tangent line, you should look for points $\left(x , y\right)$ such that $\frac{\mathrm{dy}}{\mathrm{dx}}$ does not exist.

This is due to denominator being zero for a certain point $\left(x , y\right)$.

$x - y + 5 = 0$

or

$y = x + 5$

Again, the point $\left(x , y\right)$ must lie on the parabola.

${x}^{2} - 2 x y + {y}^{2} + 6 x - 10 y + 25 = 0$

Substituting directly, we get the equation below.

${x}^{2} - 2 x \left(x + 5\right) + {\left(x + 5\right)}^{2} + 6 x - 10 \left(x + 5\right) + 25 = 0$

After expansion, we have a first order polynomial again.

$- 4 x = 0$

$x = 0$

$y = 5$

The point $\left(0 , 5\right)$ has a vertical tangent line. The equation of the tangent line is $x = 0$.

Refer to the graph below. The horizontal tangent is in cyan while the vertical tangent is in lime.