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

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 2

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

Answer 3

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)

Answer 4

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 dunno what to do after this >.>

Answer 5

Couldn't we simply do like this
y'=(-9/4)*(x/y)
As the tangent passes through (0,4)
So we plug in the values and get y'=0
that the line is parallel to x-axis

Answer 6

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

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

Answer 7

the point (0, 4) is not on the ellipse

so the tangent line will not be having a slope of 0

Answer 8

I know (0,4) is the point through which the tangent passes and y' is the slope of the tangent at point x,y . So can't we put x=0 and y=4
@TheSmartOne

Answer 9

we did, and that's how we got

(y - 4)/(x - 0) = (-9/4) * (x/y)