Answer:
(a) y = -3/5 x + 13/5
(b) y = 5/3 x + 1/3
Explanation:
(a) The slope of the tangent line is dy/dx. Use implicit differentiation:
x² + y² + 4x + 6y − 21 = 0
2x + 2y dy/dx + 4 + 6 dy/x = 0
2x + 4 + (2y + 6) dy/dx = 0
x + 2 + (y + 3) dy/dx = 0
(y + 3) dy/dx = -(x + 2)
dy/dx = -(x + 2) / (y + 3)
At the point (1, 2), the slope is:
dy/dx = -(1 + 2) / (2 + 3)
dy/dx = -3/5
Using point-slope form of a line:
y − 2 = -3/5 (x − 1)
Simplifying to slope-intercept form:
y − 2 = -3/5 x + 3/5
y = -3/5 x + 13/5
(b) The normal line is perpendicular to the tangent line, so its slope is 5/3. It also passes through the point (1, 2), so point-slope form of the line is:
y − 2 = 5/3 (x − 1)
Simplifying to slope-intercept form:
y − 2 = 5/3 x − 5/3
y = 5/3 x + 1/3