Question

Calculus 1 Question
Find the equations of both tangent lines to the graph of the ellipse x^2/4 + y^2/9 = 1 that passes through the point (0, 4) not on the graph.

Answer 1

\(\Large \frac{x^2}{4} + \frac{y^2}{9} = 1\)

Answer 2

That equation is the same thing as
9x^2 + 4y^2 = 36

And differentiating that, we get
18x + 8yy' = 0

Solving for y' we get
y' = (-9/4) * (x/y)

We know the slope is the same as the derivative
(y - 4)/(x - 0) = (-9/4) * (x/y)

(y-4)/x = (-9x)/(4y)

Cross multiply
4y^2 - 16y = -9x^2

Same thing as
9x^2 + 4y^2 = 16y

And plugging in what we know from the beginning

36 = 16y
y = 2.25

And I'm not sure what to do after this >.>

Answer 3

Maybe plug it back in to the ellipse?

Lemme try plugging that back in to the ellipse

9x^2 + 4y^2 = 36

9x^2 + 4(2.25)^2 = 36

9x^2 + 20.25 = 36

x^2 = (36 - 20.25)/9

x^2 = 1.75

\(\Large x = \pm \sqrt{1.75}\)

Answer 4

If you look at this graph
https://www.desmos.com/calculator/se3bojn9xg

How would a tangent line go through the point (0, 4) and have a slope of 0?

Answer 5

Ya I was thinking the same

Answer 6

Yeah I just graphed it myself. I think the problem has an error in it.

Answer 7

The point should be on the ellipse.

Answer 8

We need to find those two tangent lines, I believe

Answer 9

oh okay. Sorry I mis-read the question

Answer 10

when x=-1.75, 1.75 y = ???

Answer 11

It's srt(1.75) and -sqrt(1.75)

and y would equal 2.25 or -2.25

Answer 12

but we would want the positive y-value so 2.25

Answer 13

Is the eq of tangent
y=+/-1.322x+4
and the point of contact is (1.32,2.25)
and (-1.32,2.25)

Answer 14

that's what I got, thanks c:

Answer 15

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