How do you find the equations of the tangents to 5x^2-4y^2=4 at those points where the curve is cut by 5x-2y=4?

Nov 9, 2016

$5 x - 2 y = 4$

Explanation:

$5 {x}^{2} - 4 {y}^{2} = 4$ ..... [1]
$5 x - 2 y = 4$ ..... [2]

Step 1 - Find the points of intersection
If we rearrange [2] we get $2 y = 5 x - 4 \implies y = \frac{5 x - 4}{2}$

Substitute into [1]:

$5 {x}^{2} - 4 {\left(\frac{5 x - 4}{2}\right)}^{2} = 4$
$\therefore 5 {x}^{2} - 4 {\left(5 x - 4\right)}^{2} / 4 = 4$
$\therefore 5 {x}^{2} - {\left(5 x - 4\right)}^{2} = 4$
$\therefore 5 {x}^{2} - \left(25 {x}^{2} - 40 x + 16\right) = 4$
$\therefore 5 {x}^{2} - 25 {x}^{2} + 40 x - 16 = 4$
$\therefore - 20 {x}^{2} + 40 x - 20 = 0$
$\therefore 20 {x}^{2} - 40 x + 20 = 0$
$\therefore {x}^{2} - 2 x + 1 = 0$
$\therefore {\left(x - 1\right)}^{2} = 0$
$\therefore x = 1$

Using [1]; When $x = 1 \implies 5 - 2 y = 4 \implies y = \frac{1}{2}$

So there is ONE point of intersection at $\left(1 , \frac{1}{2}\right)$

Step 2 - Differentiate the equation of the curve

$5 {x}^{2} - 4 {y}^{2} = 4$

Differentiating implicitly gives:

$10 x - 8 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{10 x}{8 y} = \frac{5 x}{4 y}$ ..... [3]

Step 3 - Find the gradient of the tangent at the points of intersections
The gradient of the tangent at any particular point is given by the derivative, so using [3] at $\left(1 , \frac{1}{2}\right)$ we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(5\right) \left(1\right)}{\left(4\right) \left(\frac{1}{2}\right)} = \frac{5}{2}$

Step 4 - Find the equation of the tangent
The tangent passes through $\left(1 , \frac{1}{2}\right)$ and has gradient $m = \frac{5}{2}$.
Using $y = {y}_{1} = m \left(x - {x}_{1}\right)$, the equation is given by:

$y - \frac{1}{2} = \frac{5}{2} \left(x - 1\right)$
$\therefore 2 y - 1 = 5 \left(x - 1\right)$
$\therefore 2 y - 1 = 5 x - 5$
$\therefore 5 x - 2 y = 4$

So The line that the curve intersects with is actually also the tangent

Nov 9, 2016

graph{(4y^2-5x^2+4)(5x-2y-4)=0 [-3.96, 4.81, -1.723, 2.66]}

The given line is tangent to the curve in (1;1/2)

Explanation:

By the equation of the line $25 {x}^{2} = 4 {\left(y + 2\right)}^{2}$

Multiplying the curce equation by 5 and substituting $25 {x}^{2}$ we get

$4 {\left(y + 2\right)}^{2} - 20 {y}^{2} = 20$

Simplifying by 4 we get

${y}^{2} + 4 + 4 y - 5 {y}^{2} = 5$

$- 4 {y}^{2} + 4 y - 1 = 0$

${\left(2 y - 1\right)}^{2} = 0$

$y = \frac{1}{2}$

$x = \frac{2 y + 4}{5} = 1$

The solution found comes with multiplicity 2 so the line is tangent to the curve.