# How do you use implicit differentiation to find the points on the curve x^2+y^2=5x+4y where tangent line is horizontal and where tangent line is vertical?

Jul 7, 2017

Horizontal tagts. at (5/2,(4+sqrt41)/2), &, (5/2,(4-sqrt41)/2).

#### Explanation:

Eqn. of the Curve C is given to be, $C : {x}^{2} + {y}^{2} - 5 x - 4 y = 0.$

Diff.ing w.r.t. $x ,$ using the Chain Rule, we get,

$2 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} - 5 - 4 \frac{\mathrm{dy}}{\mathrm{dx}} = 0.$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 - 2 x}{2 \left(y - 2\right)} .$

We know that, $\frac{\mathrm{dy}}{\mathrm{dx}}$ denotes the Slope of a tangent to the

Curve at the pt.$\left(x , y\right) .$

Now, the slope of Horizontal Tangent is $0.$

$\therefore , \text{ for horizontal tgt., } \frac{\mathrm{dy}}{\mathrm{dx}} = 0 \Rightarrow \frac{5 - 2 x}{2 \left(y - 2\right)} = 0.$

$\therefore x = \frac{5}{2} , \text{ &, for the corresponding y, we appeal to "C," to get,}$

$x = \frac{5}{2} , C : \frac{25}{4} + {y}^{2} - \frac{25}{2} - 4 y = 0 , i . e . , {y}^{2} - 4 y - \frac{25}{4} = 0.$

$\therefore y = \frac{4 \pm \sqrt{16 + 25}}{2} = \frac{4 \pm \sqrt{41}}{2.}$

Thus, there are two horizontal tagts. at

(5/2,(4+sqrt41)/2), &, (5/2,(4-sqrt41)/2).

Similarly, for the Vertical Tangent, we must have undefined

$\frac{\mathrm{dy}}{\mathrm{dx}} , i . e . , y = 2 ,$ &, the corresponding $x$ can be obtained from $C .$