SOLUTION: the equations of the tangents to ellipse 9x^2+16y^2=144 from point (2,3)

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Question 823766: the equations of the tangents to ellipse 9x^2+16y^2=144 from point (2,3)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming that you are in a Calculus class. If I am wrong then you'll have to re-post (and ask for a non-Calculus solution).

9x^2+16y^2=144
To find the equation of the tangent(s), we will need the slope(s). For the slopes we will use the first derivative. And since it is a lot of work to rewrite this equation as two functions of x (one for the upper half of the ellipse and one for the lower half), we will use implicit differentiation:
2*9*x^(2-1) + 2*16*y^(2-1)*y' = 0
which simplifies to:
18x + 32y*y' = 0
Solving for y':
32y*y' = -18x
y' = %28-18x%29%2F32y
which reduces to:
y' = %28-9x%29%2F16y

Now we can use this to find the slope(s) of the tangent(s) at (2, 3):
y' = %28-9%282%29%29%2F16%283%29
which simplifies:
y' = %28-18%29%2F48
which reduces:
y' = %28-3%29%2F8
Since we found only one slope, there is only one tangent. It has a slope of -3/8 and passes through (2, 3). There are a few ways to use this data to find the equation. I like the point-slope form:
y+-+y%5B1%5D+=+m%28x-x%5B1%5D%29
Substituting our slope and the coordinates of the point we get:
y+-+%283%29+=+%28%28-3%29%2F8%29%28x-%282%29%29
Simplifying...
y+-+3+=+%28%28-3%29%2F8%29%28x-2%29
y+-+3+=+%28%28-3%29%2F8%29x-%28-3%29%2F4
y+-+3+=+%28%28-3%29%2F8%29x%2B3%2F4
This may be an acceptable solution. Or you could change it into either standard form, Ax + By = C, or slope-intercept form, y = mx + b. I'll leave it up to you to decide
  • if you need the equation in a certain form and if so...
  • which form you want and...
  • how to transform the equation we have into that form.