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?

2 Answers
Nov 9, 2016

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

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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 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.