Question

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`?

Answer 1

`5x-2y=4`

Explanation:

`5x^2-4y^2=4` ..... [1]
`5x - 2y = 4` ..... [2]

Step 1 - Find the points of intersection
If we rearrange [2] we get `2y=5x-4 => y=(5x-4)/2`

Substitute into [1]:

`5x^2-4((5x-4)/2)^2=4`
`:. 5x^2-4(5x-4)^2/4=4`
`:. 5x^2-(5x-4)^2=4`
`:. 5x^2-(25x^2-40x+16)=4`
`:. 5x^2-25x^2+40x-16=4`
`:. -20x^2+40x-20=0`
`:. 20x^2-40x+20=0`
`:. x^2-2x+1=0`
`:. (x-1)^2=0`
`:. x=1`

Using [1]; When `x=1 => 5-2y=4 => y=1/2`

So there is ONE point of intersection at `(1, 1/2)`

Step 2 - Differentiate the equation of the curve

`5x^2-4y^2=4`

Differentiating implicitly gives:

`10x-8ydy/dx=0`
`:. dy/dx = (10x)/(8y) = (5x)/(4y)` ..... [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 `(1, 1/2)` we have:

`dy/dx = ( (5)(1) ) / ( (4)(1/2) ) =5/2`

Step 4 - Find the equation of the tangent
The tangent passes through `(1, 1/2)` and has gradient `m=5/2`.
Using `y=y_1 = m(x-x_1)`, the equation is given by:

`y - 1/2 = 5/2(x-1)`
`:. 2y - 1 = 5(x-1)`
`:. 2y - 1 = 5x-5`
`:. 5x-2y=4`

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

Answer 2

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 `25x^2=4(y+2)^2`

Multiplying the curce equation by 5 and substituting `25x^2` we get

`4(y+2)^2-20y^2=20`

Simplifying by 4 we get

`y^2+4+4y-5y^2=5`

`-4y^2+4y-1=0`

`(2y-1)^2=0`

`y=1/2`

`x=(2y+4)/5=1`

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