Tangent Lines to an Ellipse

Question

Find the tangent lines to the ellipse

(E) x² + 9y² = 81

passing through the point (27,3).

Answer 1

Use implicit differentiation. Derivative is the same thing as slope.

(dy/dx) = (-x) / (9y)

This is the slope of a tangent line at any point (x, y) on the ellipse. Not the equation of the line, just the slope.

The equation of a line that goes through the point (27, 3) is equal to: (y - 3) = m(x - 27).

Solving for m, you get: m = (y - 3) / (x - 27)

Set (dy/dx) equal to m above, cross multiply, and you should get -x² +27x = 9y² - 27y

Rewrite it as 27y + 27x = x² + 9y².

The right side should look familiar as it's the original equation (x² + 9y² = 81), so...

27y + 27x = 81

Simplify.

y = -x + 3

Substitute (-x + 3) for y into the original equation, FOIL, and solve the quadratic.

You should get x = 0 and x = 54/10 as your solutions. These are the values of x, where a line tangent to the ellipse also goes through the point (3, 27).

Solve for the y-values of the points using your x-values above give you the points (0, 3) and (54/10, -24/10) as the points on the ellipse that when connected to (3, 27) are tangent to the ellipse.

Now you have two pairs of points, you can find the slope and y-intercepts and get:

y = 3

and

y = (1/4)x - 15/4

as the two equations of lines that are tangent to the ellipse x² + 9y² = 81 and also go through the point

(27, 3).

Answer 2

It's easy to see that one of the tangent lines is y = 3, but there's a second that's much more difficult to find.

Here's a procedure for finding the second tangent line:

First stage

Create a quadratic function of x that we will use to figure out slopes of the tangent lines.

All lines passing through (27,3) have the point-slope form equation: y – 3 = m(x – 27)

Solving for y we get y = mx –27m + 3

So we substitute mx –27m + 3 for y in the ellipse equation to get

x² + 9(mx –27m + 3)² = 81 which is equivalent to x² + 9(mx –27m + 3)² – 81 = 0

Expanding this collecting like terms (with the help of Symbolab) gives a quadratic equation

ax² + bx + c where the coefficients are

a = 9m² + 1, b = 54m(1 – 9m), and c = 729m(9m – 2)

Second stage

Set the discriminant of this quadratic function equal to zero, then solve for the slope m.

b² – 4ac = (54)²m²(1 – 9m)² – 4(9m² + 1)(729m(9m – 2)) = 0

= 2916m²(1 – 9m)² – 2916m(9m² + 1)(9m – 2). Here we can get divide both terms by 2916 to get

m²(1 – 9m)² – m(9m² + 1)(9m – 2) = m[m(1 – 9m)² – (9m² + 1)(9m – 2)] = 0

The two solutions are m = 0 and m = 1/4.

Third stage

Plug these two values of m (0 and 1/4) into y = mx –27m + 3 and we will have the equations of the two required tangent lines passing through the exterior point (27,3).

m = 0 gives us: y = (0)x –27(0) + 3 = 3

Producing the horizontal tangent line y = 3 (already mentioned above)

m = 1/4 gives us: y = (1/4)x –27(1/4) + 3 = (1/4)x – 27/4 +3 = (1/4)x – 15/4

Producing the second tangent line y = (1/4)x – 15/4