# Thread: Find an equation of the line tangent to the curve y1 given below that is parallel to

1. ## Find an equation of the line tangent to the curve y1 given below that is parallel to

y1 = 2x√x

y2 = 4 + 6x

y=

2. Originally Posted by plstevens
y1 = 2x√x

y2 = 4 + 6x

y=
The general equation of a line is $y - y_1 = m(x - x_1)$ where m is the gradient and $(x_1, y_1)$ is a known point on the line.

In your question, parallel to y2 = 4 + 6x means that m = 6.

To get a known point:

x-coordinate: Solve $\frac{dy_1}{dx} = 6$ for x.
y-coordinate: Substitute the value of x found above into $y_1 = 2 x \sqrt{x} = 2 x^{3/2}$.

3. ...parallel to the given line y2. thats the rest of the directions

4. Originally Posted by plstevens
...parallel to the given line y2. thats the rest of the directions
So?? I've already said that means that m = 6. Just to be crystal clear ..... it means that m = 6 in the general equation $
y - y_1 = m(x - x_1)
$
.

5. i don't get it so how does this help me to know what y1 is and what y2 is?

6. Originally Posted by mr fantastic
To get a known point: Mr F adds: That is, to get the $(x_1 , y_1)$

x-coordinate: Solve $\frac{dy_1}{dx} = 6$ for x.
y-coordinate: Substitute the value of x found above into $y_1 = 2 x \sqrt{x} = 2 x^{3/2}$.
Originally Posted by plstevens
i don't get it so how does this help me to know what y1 is and what y2 is?
x-coordinate: Solve $\frac{dy_1}{dx} = 6$ for x:

$\frac{dy_1}{dx} = \frac{3}{2} \times 2x^{\frac{3}{2} - 1} = 3 x^{1/2}$.

So solve $3 x^{1/2} = 6$:

$\Rightarrow x^{1/2} = 2$

$\Rightarrow x = 4$.

Now substitute x = 4 into $y_1 = 2 x^{3/2} = 2 x x^{1/2} = 2 x \sqrt{x}$:

$y = 2 (4) \sqrt{4} = 2 (4) (2) = 16$.

So a known point is (4, 16) ........

So $x_1 = 4$ and $y_1 = 16$ .......