Find the equations of both of the tangent lines to the ellipse x^2+4y^2=36 that pass through the point (12,3).

Find the equations of both of the tangent lines to the ellipse x^2+4y^2=36 that pass through the point (12,3). List the line with the smaller slope first

thank you!!

Hello,

x² + 4y² = 36
y² = 9 - x²/4
y = ±√(9 - x²/4)

Let's study
f(x) = √(9 - x²/4)    and    g(x) = -√(9 - x²/4)
f'(x) = -¼.x / √(9 - x²/4)    and     g'(x) = ¼.x / √(9 - x²/4)

A tangent line to the ellipse at point x₀ would have as equation:
y = f'(x₀).(x - x₀) + f(x₀)

If that tangent line passes through (12; 3), thosee coordinates fits its equation:
3 = f'(x₀).(12 - x₀) + f(x₀)
3 = -¼.x₀(12 - x₀) / √(9 - x₀²/4) + √(9 - x₀²/4)
6√(36 - x₀²) = -x₀(12 - x₀) + 36 - x₀²
6√(36 - x₀²) = -12x₀ + 36
√(36 - x₀²) = -2x₀ + 6
36 - x₀² = 4x₀² - 24x₀ + 36
5x₀² - 24x₀ = 0
x₀=0 or x₀=24/5
Solution x₀=24/5 must be rejected (because -2x₀+6 must be positive).

So the equation is:
y = f'(x₀).(x - x₀) + f(x₀)
y = f'(0).(x - 0) + f(0)
y = 0.x + √9
y = 3

Repeat the process with g instead of f and you shall get the other equation. :-)

Methodically,
Dragon.Jade :-)

Do your own homework. -____________-