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.
Am I on the right track with this?
Thank you for any help
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
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
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?
Then the gradient of the tangent is .
But since the tangent passes through the point (12, 3) the gradient is also given by .
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.