# Thread: A line tangent to an implicit function, and through two points not on the function

1. ## A line tangent to an implicit function, and through two points not on the function

I have been given the implicit function: X^2+4y^2=36, and have been asked to find a line tangent to it through the point (12,3), which is not on the curve.

I know there is one horizontal tangent line, at y=3. However, I am having trouble finding the non=horizontal line.

I know that the derivative is -x/4y. That is fairly simple. I think that I have to set the equation of the line through (12,3) equal to the implicit function, and solve for the slope (which is the derivative) in order to get a numerical value. But I am lost as to how to do that exactly.

(-x/4y)(x-120+3=x^2+4y^2-36

Am I on the right track with this?

Thank you for any help
Respectfully
Physics Student

2. We know that the slope of such a line would be

$\displaystyle \frac{3-y}{12-x}=y'(x)=-\frac{x}{4y},$

where $\displaystyle (x,y)$ is a desired point on the curve. Now we may solve for $\displaystyle x$ in terms of $\displaystyle y$ to find the points.

3. Originally Posted by Scott H
We know that the slope of such a line would be

$\displaystyle \frac{3-y}{12-x}=y'(x)=-\frac{x}{4y},$

where $\displaystyle (x,y)$ is a desired point on the curve. Now we may solve for $\displaystyle x$ in terms of $\displaystyle y$ to find the points.
I'm still confused. I can solve for x in terms of y. I know that the equation is going to be

$\displaystyle \frac{-x}{4y}*(x-12)+3$

Using what you gave me, I get y= two rather complicated functions. Should I then plug both of those into the originall implicit and solve for x, then y?

Thanks for the response
Physics Student

4. ## My Attempt

When you rearrange for y, you get y=sqrt[9- (1/4)x^2]

so as as said before, your slope is

(3-y)/(12-x) which should equal your derivative -x/4y

(3-y)/(12-x) = -x/4y

when you replace your y with sqrt[9- (1/4)x^2] and simplify you should get

x^2-11x+24 =0

use quadratic to get x= [11 +/- sqrt(5) ]/2

That's what I got at least. Although I've been struggling with these problems recently. Does the back of your book give this answer if any?

5. I don't have any solution to the problem that I can check my work against, sorry.

6. This is homework I assume. Could you do me a favor and post back with the solution. I'm pretty sure my method was correct, It's always worked for me except for one question (in the thread I JUST started). I really need the mental boost.

7. Originally Posted by PhysicsStudent
I have been given the implicit function: X^2+4y^2=36, and have been asked to find a line tangent to it through the point (12,3), which is not on the curve.

I know there is one horizontal tangent line, at y=3. However, I am having trouble finding the non=horizontal line.

I know that the derivative is -x/4y. That is fairly simple. I think that I have to set the equation of the line through (12,3) equal to the implicit function, and solve for the slope (which is the derivative) in order to get a numerical value. But I am lost as to how to do that exactly.

(-x/4y)(x-120+3=x^2+4y^2-36

Am I on the right track with this?

Thank you for any help
Respectfully
Physics Student
Let the required tangent be tangent to the ellipse at the point $\displaystyle \left(2 \sqrt{9 - a^2}, -a \right)$ where a > 0 (a simple diagram shows why this is the case).

Then the gradient of the tangent is $\displaystyle m = \frac{\sqrt{9 - a^2}}{2a}$.

But since the tangent passes through the point (12, 3) the gradient is also given by $\displaystyle m = \frac{3 + a}{12 - 2 \sqrt{9 - a^2}}$.

Therefore $\displaystyle \frac{\sqrt{9 - a^2}}{2a} = \frac{3 + a}{12 - 2 \sqrt{9 - a^2}}$.

Solve this equation for $\displaystyle a$ (remember you want a > 0). I get $\displaystyle a = \frac{9}{5}$.

So the required tangent is tangent to the ellipse at the point $\displaystyle \left( \frac{24}{5}, \frac{9}{5}\right)$.

It should now be simple to get the equation of the required tangent.