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
Explanation:
5x^2-4y^2=4 ..... [1]
5x - 2y = 4 ..... [2]
Step 1 - Find the points of intersection
If we rearrange [2] we get
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
So there is ONE point of intersection at
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
dy/dx = ( (5)(1) ) / ( (4)(1/2) ) =5/2
Step 4 - Find the equation of the tangent
The tangent passes through
Using
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
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
Explanation:
By the equation of the line
Multiplying the curce equation by 5 and substituting
Simplifying by 4 we get
The solution found comes with multiplicity 2 so the line is tangent to the curve.