How do I find 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)

Question

The answer is y=4x/3 + 4 and y= -4x/3 + 4; but when I solve for the dy/dx I get the slope is = 0 so there is only one equation y=4; However when I graphed this and it is not tangent to the eclipse.

I don't know why or how to solve this equation, my class just went on break and don't know where i went wrong.

Thank you.

Answer 1

We start by identifying the relationship between the ellipse and the unit circle, since points of tangency and tangent lines are much easier to find for the unit circle.

We get the ellipse (x²/4) + (y²/9) = 1

by dilating the unit circle x² + y² = 1 with scale factors k_x = 2 and k_y = 3

The tangent lines from the point (0, 4/3) to the unit circle pass through

points of tangency (√7/4, 3/4) and (–√7/4, 3/4)

Slopes of these tangent lines are

–√7/3 passing through (√7/4, 3/4) and

√7/3 passing through (–√7/4, 3/4)

For the ellipse the points of tangency are

those obtained for the unit circle and multiplying

x-coordinates by k_x = 2 and

y-coordinates by k_y = 3

Slopes for the tangent lines are obtained by taking

those obtained for the unit circle and multiplying by k_y/k_x

(–√7/3)(k_y/k_x) = –√7/2 passing through (√7/2, 9/4) and

(√7/3) (k_y/k_x) = √7/2 passing through (–√7/2, 9/4)

Equations for the tangent lines to the ellipse passing through (0,4) are

y = (–√7/2)(x – √7/2) + 9/4 = (–√7/2)x + 4

y = (√7/2)(x + √7/2) + 9/4 = (√7/2)x + 4

Desmos graph: desmos.com/calculator/rb8hrmkoas

Answer 2

maybe you miscopied the point, as (0,4) isn't on the ellipse, (0,3) is

tangent line at (0,4) does not exist

tangent at (0,3) has slope 0

but the problem is solvable if it means the tangent lines pass through the point (0,4) while (0,4) is not a point on the ellipse

since the tangent lines are symmetric about the y axis, the equation of the lines in slope intercept form have a y intercept = 4 as in (0,4)

y =mx +4, now just solve for m = slope, which will be plus or negative some number, about |1| from looking at the graph

about = +/-1.328

y= +/-1.328x + 4

1.328 is close to 4/3= 1.333...

rounded off to nearest 10th, they are the same