Question

Write the equation of the tangent line to the curve x^2/8 - y^2/4 =1 at the point (4,2) by using the following facts. The slope m of the tangent line to a hyperbola at the point (x, y) is: m=b^2x/a^2y for x^2/a^2 -y^2/b^2=1 m=a^2x/b^2y for y^2/a^2 - x^2/b^2 =1

Answer 1

Answer: y = x - 2

Explanation:

First you take the derivative of each term. d/dx(x²/8) - d/dx(y²/4) = d/dx(1)

x/4 - (y/2)dy/dx = 0

Then you solve for dy/dx: dy/dx = x/2y

Plug in the values: dy/dx = 1

To find the y-intercept, plug in values for y = mx+ b. 2 = 4 + b, b = -2

The equation is y = x - 2

Answer 2

`y=x-2`

Explanation:

So we are given the formula for the slope of a hyperbola in this form:

`(x^2)/(a^2)-(y^2)/(b^2)=1`.

That formula for the slope is
`m=(b^2x)/(a^2y)`

If we compare the following two equations, we will be able to find
`a^2` and
`b^2`:

`(x^2)/(a^2)-(y^2)/(b^2)=1`

`(x^2)/(8)-(y^2)/(4)=1`

We see that
`a^2=8` while
`b^2=4`.

So the slope at
`(x,y)=(4,2)` is:

`m=(b^2x)/(a^2y)=(4(4))/(8(2))=(16)/(16)=1`.

Recall: Slope-intercept form of a linear equation is
`y=mx+b`.

We just found
`m=1`. Let's plug that in.

`y=1x+b`

`y=x+b`

To find
`b`, the
`y`-intercept, we will need to use a point on our tangent line. We know that it is going through
`(4,2)`.

Let's enter this point in to find
`b`.

`2=4+b`

Subtract 4 on both sides:

`2-4=b`

Simplify:

`-2=b`

The equation for the tangent line at
`(4,2)` on the given equation is:

`y=x-2`