tangent question (1 Viewer)

[ephemeral]

this one seems pretty easy, but im most likely too stupid to think.

A tangent to the hyperbola y=1/(2x), x>0 cuts the y-axis at the point y=2.
Find the gradient of this tangent.

 

[Jago]

at y = 2, 2x = 1/2, x = 1/4

dy/dx = -1/(2x²) = 0 at x = 1/4

m = -8

 

[ephemeral]

I did something like that as well, but the answer says the gradient is -2.

 

[Trev]

Well the answer is wrong.

 

[Riviet]

-And Jago is correct.

 

[speed2]

at y = 2, 2x = 1/2, x = 1/4

dy/dx = -1/(2x²) = 0 at x = 1/4

m = -8

but it doesn't say that the tangent lies on the hyperbole at y=2

 

[Antwan23q]

ahhh the plot thickens! so the point is (0,2)

 

[Jago]

yeah, tangent at y = 2. but at y = 2, x = 1/4 (by subbing y=2 into y=1/(2x))

anyway it can't be (0,2) because the line is a hyperbola

 

[Antwan23q]

no, the tangent isnt at y=2, the tangent is at some point, and then it cuts the y axis a y=2

 

[Antwan23q]

perhaps someone should graph it

 

[Jago]

oh, misread the question.

 

[Antwan23q]

yes, but i dont think it can be solved if u dont know the point on the hyperbola

 

[Bokky]

isnt the derivative of y = 1/(2x) dy/dx = -2/x^2 ?

 

[Trev]

isnt the derivative of y = 1/(2x) dy/dx = -2/x^2 ?

No, it isn't.

 

[acmilan]

A tangent to the hyperbola y=1/(2x), x>0 cuts the y-axis at the point y=2.
Find the gradient of this tangent.

Things we know:

Point on the tangent: (0,2)

Gradient of tangent: m = -1/(2x₁²), where x₁ is the point that the tangent cuts the hyperbola.

Equation of tangent: y - 2 = m(x - 0) -> y = mx + 2

Equate equations:

mx + 2 = 1/(2x)
2mx² + 4x - 1 = 0

Since its a tangent, theres only one root, or two equal roots, so delta = 0

b² - 4ac = 0
16 + 8m = 0
2 + m = 0
m = -2

So the gradient is -2.

When m = -2, x₁ = 1/2, y₁ = 1

So the tangent of the hyperbola y = 1/2x that cuts the y-axis at the point y=2 is y = -2x + 2 and touches the hyperbola at (0.5,1)

 

[speed2]

acmilan ur a bloody GENIUS!

 

[king_of_boredom]

A tangent to the hyperbola y=1/(2x), x>0 cuts the y-axis at the point y=2.
Find the gradient of this tangent.

Things we know:

Point on the tangent: (0,2)

Gradient of tangent: m = -1/(2x₁²), where x₁ is the point that the tangent cuts the hyperbola.

Equation of tangent: y - 2 = m(x - 0) -> y = mx + 2

Equate equations:

mx + 2 = 1/(2x)
2mx² + 4x - 1 = 0

Since its a tangent, theres only one root, or two equal roots, so delta = 0

b² - 4ac = 0
16 + 8m = 0
2 + m = 0
m = -2

So the gradient is -2.

When m = -2, x₁ = 1/2, y₁ = 1

So the tangent of the hyperbola y = 1/2x that cuts the y-axis at the point y=2 is y = -2x + 2 and touches the hyperbola at (0.5,1)

without really reading it, it looks right. anyway heres a little something that might help with diagrammatic representation:

 

[Meads]

that question is screwed...lol

 

[Bokky]

how is dy/dx = -2/x^2 not the derivative of y = 1/(2x)

1/(2x) is the same as 2x^-1 isnt it, u times the front by -1 and u bring the power down by 1 giving u x^-2 so that gives u -2x^-2 therefore -2/(x^2) ????

 

[acmilan]

No, 1/(2x) = (1/2)x^-1 = 2^-1x^-1