y = -¼ (x − 1) + 4
Step-by-step explanation:
The easiest way is with geometry.
y = √(17 − x²) can be rewritten as x² + y²= 17, which is a circle with center (0, 0) and radius √17.
The line perpendicular to the tangent line at the point of intersection must pass through the center of the circle. In other words, the points (1, 4) and (0, 0) must both be points on the perpendicular line. So the slope of the tangent line is:
-1 / m = (4 − 0) / (1 − 0)
-1 / m = 4
m = -¼
So the equation of the tangent line is y = -¼ (x − 1) + 4.
Another way is with algebra.
The line passes through the point (1, 4), so we can write it as:
y = m(x − 1) + 4
y = mx − m + 4
y = mx − (m − 4)
The intersection of the line and the curve is:
mx − (m − 4) = √(17 − x²)
m² x² − 2m (m − 4) x + (m − 4)² = 17 − x²
(m² + 1) x² − 2m (m − 4) x + (m − 4)² − 17 = 0
(m² + 1) x² − 2m (m − 4) x + m² − 8m + 16 − 17 = 0
(m² + 1) x² − 2m (m − 4) x + (m² − 8m − 1) = 0
Since the line is tangent to the curve, they intersect at only one point, so this equation can only have one solution. That means the discriminant of the quadratic is equal to 0.
b² − 4ac = 0
(-2m (m − 4))² − 4 (m² + 1) (m² − 8m − 1) = 0
4m² (m − 4)² − 4 (m² + 1) (m² − 8m − 1) = 0
m² (m − 4)² − (m² + 1) (m² − 8m − 1) = 0
m² (m² − 8m + 16) − (m² + 1) (m² − 8m − 1) = 0
m² (m² − 8m + 16) = (m² + 1) (m² − 8m − 1)
m⁴ − 8m³ + 16m² = m² (m² − 8m − 1) + (m² − 8m − 1)
m⁴ − 8m³ + 16m² = m⁴ − 8m³ − m² + m² − 8m − 1
16m² = -8m − 1
16m² + 8m + 1 = 0
(4m + 1)² = 0
4m + 1 = 0
m = -¼
The line is therefore y = -¼ (x − 1) + 4.
