# Question #a51c8

##### 1 Answer

The point

The point

#### Explanation:

Firstly, here are the steps for the differentiation.

Differentiate everything (implicitly) with respect to

#2x - 2 (y + x frac{dy}{dx}) + 2y frac{dy}{dx} + 6 - 10 frac{dy}{dx} = 0#

Bring all the terms containing

#2x - 2y + 6 = 2x frac{dy}{dx} - 2y frac{dy}{dx} + 10 frac{dy}{dx}#

Remove the common multiple of 2.

#x - y + 3 = x frac{dy}{dx} - y frac{dy}{dx} + 5 frac{dy}{dx}#

Make

#frac{dy}{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

#frac{dy}{dx} = frac{ x - y + 3 }{x - y + 5} = 0#

which simplifies to

#y = x + 3#

Bear in mind that the point

#x^2 - 2xy + y^2 + 6x - 10y + 25 = 0#

Solve this simultaneous equation by substitution.

#x^2 - 2x(x + 3) + (x + 3)^2 + 6x - 10(x + 3) + 25 = 0#

Expanding the terms reveal that the

#4 - 4x = 0#

#x = 1#

#y = 1 + 3 = 4#

**The point #(1,4)# 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

This is due to denominator being zero for a certain point

#x - y + 5 = 0#

or

#y = x + 5#

Again, the point

#x^2 - 2xy + y^2 + 6x - 10y + 25 = 0#

Substituting directly, we get the equation below.

#x^2 - 2x(x + 5) + (x + 5)^2 + 6x - 10(x + 5) + 25 = 0#

After expansion, we have a first order polynomial again.

#-4x = 0#

#x = 0#

#y = 5#

**The point #(0,5)# 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.