Using calculus would make things easier, but I will assume you know
only algebra.
We use two facts:
1. A line is tangent to a curve if the system consisting
of the equations of the line and the conic section has exactly one
solution.
2. A quadratic equation has exactly one solution if the
discriminant B²-4AC = 0
Let the slope of the tangent line be m.
Since the line passes through (-1,-7), we use the point-slope form:
y-y1 = m(x-x1)
y-(-7) = m(x-(-1))
y+7 = m(x+1)
y+7 = mx+m
y = mx+m-7 <---(The equation of the tangent line)
We know that if we solve this system
we would get (x,y) = the point of tangency,
So we solve that by substitution:
x² - (mx+m-7)² = 16
x² - (m²x²+m²+49+2m²x-14mx-14m) = 16
x² - m²x² - m² - 49 - 2m²x + 14mx + 14m = 16
x² - m²x² - 2m²x + 14mx - m² - 49 + 14m = 16
(1-m²)x² + (-2m²+14m)x + (-m²+14m-65) = 0
This must have a single solution, so its discriminant
B²-4AC = 0
Discriminant = (-2m²+14m)² - 4(1-m²)(-m²+14m-65) =
(4m4-56m³+196m²) - (4-4m²)(-m²+14m-65) =
(4m4-56m³+196m²) - (-4m²+56m-260+4m4-56m³+260m²) =
4m4 - 56m³ + 196m² + 4m² - 56m + 260 - 4m4 + 56m³ - 260m² =
-60m² - 56m + 260
This must = 0 so that the line will be tangent to the
hyperbola:
-60m² - 56m + 260 = 0
Divide through by -4
15m² + 14m - 65 = 0
(3m-5)(5m+13) = 0
3m-5 = 0; 5m+13 = 0
3m = 5; 5m = -13
m = ; m =
So we have two solutions:
using m =
y = mx+m-7 becomes
y = x + - 7
Multiply through by 3
3y = 5x + 5 - 21
3y = 5x - 16
-5x + 3y = -16
5x - 3y = 16
using m =
y = mx+m-7 becomes
y = x + - 7
Multiply through by 3
5y = -13x - 13 - 35
5y = -13x - 48
13x + 5y = -48
So there are two tangent lines. Their equations are
5x - 3y = 16 and 13x + 5y = -48.
Drawing in the other one:
Edwin
