Find all tangent lines to the ellipse x^2+4y^2=5 which pass through th point (-5,0).?

Question

Your solution should describe how you know that you have found all of the solutions.

Answer 1

Find all tangent lines to the ellipse x² + 4y² = 5 which pass through the point P(-5,0).

A cursory look at a graph show there can only be two such tangent lines. And they are symmetrical about the x-axis.

At the point of tangency the tangent line and the ellipse will have the same slope. Take the derivative to find the slope of the ellipse.

x² + 4y² = 5
4y² = 5 - x²
y² = (5 - x²)/4
y = ±√(5 - x²) / 2

dy/dx = ± x / [2√(5 - x²)]

Let's focus on the positive slope. Once we find that tangent line, the other will be symmetrical.

dy/dx = -x / [2√(5 - x²)]

Find the slope of the tangent line from P(-5,0) to the point of tangency Q(x,y) on the ellipse.

Q(x,y) = Q(x, √(5 - x²) / 2)

Slope m = ∆y/∆x = (√(5 - x²)/2 - 0) / (x + 5)
m = √(5 - x²) / [2(x + 5)]

Set the two slopes equal and solve for x.

m = √(5 - x²) / [2(x + 5)] = -x / [2√(5 - x²)]
2(5 - x²) = -2x(x + 5)
(5 - x²) = -x(x + 5)
5 - x² = -x² - 5x
5 = -5x
x = -1

The slope of the tangent line is:

m = -x / [2√(5 - x²)] = 1 / [2√(5 - 1)] = 1/4

With the slope m and the point P(-5,0) we can write the equation of the tangent line.

y - 0 = (1/4)(x + 5)
y = (1/4)(x + 5)
y = x/4 + 5/4

The other tangent line is:

y = (-1/4)(x + 5)
y = -x/4 - 5/4
__________

The points of tangency are (-1, 1) and (-1, -1).