# How do you find the equation of the tangent lines to the curve at the point x=2 considering the equation x^2-2xy+4y^2=64?

Nov 7, 2016

$244 y - \left(61 + 3 \sqrt{61}\right) x + 128 \sqrt{61} = 0$ and
$244 y - \left(61 - 3 \sqrt{61}\right) x - 128 \sqrt{61} = 0$

#### Explanation:

To find the equation of the tangent we need to find the gradient of the tangent so we need the derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$

I will assume you know hoe differentiate implicitly and to apply the product rule. If not there are many examples to look through. I will do each part in a separate step so hopefully you can follow

We have: ${x}^{2} - 2 x y + 4 {y}^{2} = 64$

Differentiating wrt $x$ we have

$\frac{d}{\mathrm{dx}} \left({x}^{2}\right) - \frac{d}{\mathrm{dx}} \left(2 x y\right) + \frac{d}{\mathrm{dx}} \left(4 {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(64\right)$

The first and last term we can just deal with using straight forward rules to give:

$2 x - \frac{d}{\mathrm{dx}} \left(2 x y\right) + \frac{d}{\mathrm{dx}} \left(4 {y}^{2}\right) = 0$

For the next term we apply the product rule to $\left(2 x\right) \left(y\right)$ give:

$2 x - \left\{\left(2 x\right) \frac{d}{\mathrm{dx}} \left(y\right) + \left(y\right) \frac{d}{\mathrm{dx}} \left(2 x\right)\right\} + \frac{d}{\mathrm{dx}} \left(4 {y}^{2}\right) = 0$
$\therefore 2 x - \left\{\left(2 x\right) \frac{\mathrm{dy}}{\mathrm{dx}} + \left(y\right) \left(2\right)\right\} + \frac{d}{\mathrm{dx}} \left(4 {y}^{2}\right) = 0$
$\therefore 2 x - 2 x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y + \frac{d}{\mathrm{dx}} \left(4 {y}^{2}\right) = 0$

And for the final term we need to differentiate implicit (by applying the chain rule) to give:
$2 x - 2 x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y + \frac{d}{\mathrm{dy}} \left(4 {y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore 2 x - 2 x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y + 8 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore x - x \frac{\mathrm{dy}}{\mathrm{dx}} - y + 4 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

We can now factorise $\frac{\mathrm{dy}}{\mathrm{dx}}$ to get an explicit function for the derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(4 y - x\right) + x - y = 0$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \left(4 y - x\right) = y - x$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y - x}{4 y - x}$

We want the tangent when $x = 2$

To get the coordinates on the curve where $x = 2$ we substitute $x = 2$ into the equation ${x}^{2} - 2 x y + 4 {y}^{2} = 64$:

$x = 2 \implies {2}^{2} - 2 \left(2\right) y + 4 {y}^{2} = 64$
$\therefore 4 - 4 y + 4 {y}^{2} = 64$
$\therefore 4 {y}^{2} - 4 y - 60 = 0$
$\therefore {y}^{2} - y - 15 = 0$

This does not factorise, so we can solve by completing the square (or using the quadratic formula):
${\left(y - \frac{1}{2}\right)}^{2} - {\left(\frac{1}{2}\right)}^{2} - 15 = 0$
${\left(y - \frac{1}{2}\right)}^{2} - \frac{1}{4} - 15 = 0$
${\left(y - \frac{1}{2}\right)}^{2} - \frac{1}{4} - 15 = 0$
${\left(y - \frac{1}{2}\right)}^{2} = \frac{61}{4}$
$y - \frac{1}{2} = \pm \frac{\sqrt{61}}{2}$
$y = \frac{1}{2} \pm \frac{\sqrt{61}}{2}$

So the two points we are interested in are
$A \left(2 , \frac{1 - \sqrt{61}}{2}\right)$ and $B \left(2 , \frac{1 + \sqrt{61}}{2}\right)$

Let us consider A first:
At $A \left(2 , \frac{1 - \sqrt{61}}{2}\right)$,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{1 - \sqrt{61}}{2} - 2}{4 \frac{1 - \sqrt{61}}{2} - 2}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{1 - \sqrt{61}}{2} - 2}{2 - 2 \sqrt{61} - 2}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{- 3 - \sqrt{61}}{2}}{- 2 \sqrt{61}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 + \sqrt{61}}{4 \sqrt{61}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 + \sqrt{61}}{4 \sqrt{61}} \cdot \frac{\sqrt{61}}{\sqrt{61}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 \sqrt{61} + 61}{4 \cdot 61}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{61 + 3 \sqrt{61}}{244}$

So the tangent has equation
$y - \frac{1 - \sqrt{61}}{2} = \frac{61 + 3 \sqrt{61}}{244} \left(x - 2\right)$
$\therefore 244 y - 122 \left(1 - \sqrt{61}\right) = \left(61 + 3 \sqrt{61}\right) \left(x - 2\right)$
$\therefore 244 y - 122 + 122 \sqrt{61} = 61 x - 122 + 3 \sqrt{61} x - 6 \sqrt{61}$
$\therefore 244 y - \left(61 + 3 \sqrt{61}\right) x + 128 \sqrt{61} = 0$

And at $B \left(2 , \frac{1 + \sqrt{61}}{2}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{1 + \sqrt{61}}{2} - 2}{4 \frac{1 + \sqrt{61}}{2} - 2}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{1 + \sqrt{61}}{2} - 2}{2 + 2 \sqrt{61} - 2}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{- 3 + \sqrt{61}}{2}}{2 \sqrt{61}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 + \sqrt{61}}{4 \sqrt{61}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 + \sqrt{61}}{4 \sqrt{61}} \cdot \frac{\sqrt{61}}{\sqrt{61}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 \sqrt{61} + 61}{4 \cdot 61}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{61 - 3 \sqrt{61}}{244}$

So the tangent has equation
$y - \frac{1 + \sqrt{61}}{2} = \frac{61 - 3 \sqrt{61}}{244} \left(x - 2\right)$
$\therefore 244 y - 122 \left(1 + \sqrt{61}\right) = \left(61 - 3 \sqrt{61}\right) \left(x - 2\right)$
$\therefore 244 y - 122 - 122 \sqrt{61} = 61 x - 122 - 3 \sqrt{61} x + 6 \sqrt{61}$
$\therefore 244 y - \left(61 - 3 \sqrt{61}\right) x - 128 \sqrt{61} = 0$