1. ## More Parametrics

Find the equations of the tangents to the given curve that pass through the point (15, 15).

$x = 9t^2 + 6$
$y = 6t^3 + 9$

I isolated t from x:

$t=\pm \sqrt {\frac {x-6}{9}}$

Plugged it into the y equation:

$y = \pm 6\left(\sqrt {\frac {x-6}{9}}\right)^3 + 9$

$y = \pm 162\left(\sqrt {x-6}\right)^3 + 9$

$y = \pm 162(x-6)^{\frac {3}{2}} + 9$

Then found the the derivatives of y:

$y'=\pm 243\sqrt {x-6}$

Plugged in when x and y are equal to 15:

$15=\pm 243\sqrt {15-6}\cdot 15 +b$

$
b = 10950, 10920$

So I found my two equation to be:

$
y=243\sqrt {x-6} + 10920$

and

$y=-243\sqrt {x-6} + 10950$

But these aren't the right answers. Any suggestions?

2. Originally Posted by topher0805
Find the equations of the tangents to the given curve that pass through the point (15, 15).

$x = 9t^2 + 6$
$y = 6t^3 + 9$

I isolated t from x:

$t=\pm \sqrt {\frac {x-6}{9}}$

Plugged it into the y equation:

$y = \pm 6\left(\sqrt {\frac {x-6}{9}}\right)^3 + 9$

...
From this equation you get:

$y = 9 \pm \frac29 (x-6)^{\frac32}$

Then the first derivative is:

$y' = \pm \frac13 \sqrt{x-6}$

I'll leave the rest for you.

3. Oh, I was multiplying 6 by 27 when I should have been dividing 6 by 27.

Wow, I'm an idiot tonight.

4. $t = \pm \sqrt{\frac{x - 6}{9}}$

Bit of a calculation error, after substituting in:
$y = \pm 6 \left(\sqrt{\frac{x - 6}{9}}\right)^{3} + 9$
$y = \pm 6 \left({\color{blue}\frac{1}{3}} \sqrt{x - 6}\right)^{3} + 9$
$y = \pm \frac{6}{27} \left(\sqrt{x-6}\right)^{3} + 9$

----

Why don't you just use differentiate it with respect to t:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\left(6t^{3} + 9\right)'}{\left(9t^{2} + 6\right)'} = \frac{18t^{2}}{18t} = t$

You can easily find t by solving for it since you know x = 15 and y = 15.

Edit: Jee I'm slow lol.

5. Originally Posted by earboth
From this equation you get:

$y = 9 \pm \frac29 (x-6)^{\frac32}$

Then the first derivative is:

$y' = \pm \frac13 \sqrt{x-6}$

I'll leave the rest for you.
So my two equations would then be:

$
y=x-10$

and

$
y=-x+40$

Is this correct?

6. Are those your tangent equations? They should be in the form of y = mx + b ...

7. Originally Posted by o_O
Are those your tangent equations? They should be in the form of y = mx + b ...
Sorry, is that right now?

8. You can simplify that a great deal although the point (15,15) doesn't satisfy any of the two equations.

9. Here's what I did:

I knew that the derivative is:

$
y=\pm \frac {\sqrt {x-6}}{3}$

Then I used that and the point (15,15) to find b.

$15 = \frac {\sqrt {15-6}}{3}\cdot 15 + b$

$
b = 0$

So I have one equation:

$y = x
$

And the other one is:

$
y = -x +30$

Oh I see.......I'm going crazy................

10. Just to add more to this. $y = -x + 30$ does not work.

Why?

You derived that by assuming t = -1 (one of your two solutions to t at the beginning). However, if you plug t = -1 to your initial parametric equation for y:
$15 = 6t^{3} + 9$
$15 = 6(-1)^{3} + 9$
$15 \neq 3$

So only one tangent works

11. I just wanted to revive this, being that I have a similar question on my homework. There are two tangents, and I can't find the answer anywhere on the internet. The key factor is that the tangent to the curve only has to pass through the point listed, not be tangent to the curve at that point.

For this equation, which looks like "<" there will be two tangents that pass through that point.

If clarification is needed, you can consult a graphing calculator to see that two DIFFERENT tangents to the curve will pass through the point listed. I need to find both equations.

Thanks