We know that the slope of such a line would be
where is a desired point on the curve. Now we may solve for in terms of to find the points.
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
I'm still confused. I can solve for x in terms of y. I know that the equation is going to be
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
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?
Let the required tangent be tangent to the ellipse at the point where a > 0 (a simple diagram shows why this is the case).
Then the gradient of the tangent is .
But since the tangent passes through the point (12, 3) the gradient is also given by .
Therefore .
Solve this equation for (remember you want a > 0). I get .
So the required tangent is tangent to the ellipse at the point .
It should now be simple to get the equation of the required tangent.