Answer to Question #129232 in Analytic Geometry for Sym

Question

Determine the equation of the tangents at the end of the latus rectum of the parabola with the equation
x²+4x-8y-4=0

Answer 1

1) Vertex of parabola is at point (-2, -1):

x² + 4x - 8y -4 = 0

(x + 2)² - 8 = 8y

(x + 2)² = 8(y + 1)

(x - (-2))= 8(y - (-1)).

Notice, that latus rectum (line that crosses the focus of parabola) is parallel to x-axis.

2) One more time:

x² + 4x - 8y -4 = 0

(x + 2)² - 8 = 8y

(x + 2)² = 8(y + 1)

Using the modified formula for parabola (x)² = 2p(y),

2p = 8 <=> p = 4.

Focus is generally at point (0, (p/2)), but as we have specific parametes because of the vertex, the formula of focus becomes (-2, (p/2) - 1), so the focus will be at:

(-2, (4/2) - 1) = (-2, 2 - 1) = (-2, 1).

3) Using the y-coordinate of our focus,

(x + 2)² = 8(1 + 1)

(x + 2)² = 16

x + 2 = ±4

x = -6 or x = 2.

As the latus rectum is parallel to x-axis, we get two endpoints on the graph of parabola:

(-6, 1) and (2, 1).

4) Let's find the derivative of equation of the parabola:

(x² + 4x - 8y - 4)' = 0

2x + 4 - 8yy' = 0

x + 2 - 4yy' = 0

4yy' = - x - 2

y' = (-x-2) / (4y)

Putting the coordinates of the two points, we get two version of derivatives:

y₁' = (-(-6)-2) / (4*1) = 4 / 4 = 1 at (-6, 1)

y₂' = (-2-2) / (4*1) = -4 / 4 = -1 at (2, 1)

5) Finally, we write the equations of tangents using the general formula

y = y'(x₀)(x - x₀) + y₀ at (x₀, y₀):

y₁ = -1(x - (-6)) + 1 = -x + 2 + 1 = -x + 3 at (-6, 1)

y₂ = 1(x - 2) + 1 = x + 6 + 1 = x + 7 at (2, 1).

Answer: y₁ = -x + 3, y₂ = x + 7.