Geometric Applications Problem (Derivatives)

Question

Find equation of the tangent and normal line to the curve x^2 + y^2 + 4x + 6y + 5 = 0 at (x,y) where x < 0 and y < 0 , Given that the tangent line at this point makes an angle of measure 45 with the positive direction of x-axis. 

Answer 1

The equation of curve is x² + y² + 4x + 6y + 5 = 0 at (x,y) where x < 0 and y < 0.

Your description is lacking or in error.

The equation is a circle with center point at (- 2, - 3) and a radius of √5.

The tangent line at 45 degrees intersection with the +x axis is y = x - 5.

Here the y intercept is - 5 but x is 0, not negative as you specified.

There is no tangent line with x < 0 and y < 0 at a 45 deg angle to +x.

To discover this use implicit differentiation

y² + 6y = - x² - 4x - 5

Apply derivative with respect to x.

2y dy/dx + 6 dy/dx = - 2x - 4.

(2y + 6) dy/dx = - 2x - 4.

dy/dx = (- 2x - 4)/(2y + 6).

dy/dx = (- x - 2)/(y + 3).

The tangent line at this point (x, y) makes an angle of measure 45 with the positive direction of x -axis.

The slope of tangent line m = tan(θ) = tan(45^o) = 1.

To find the equation of the tangent line, dy/dx = 1 → (- x - 2)/(y + 3) = 1.

To find the y intercept, substitute the value of x = 0 in the above equation.

[- (0) - 2 ]/[y + 3] = 1

(- 2)/[y + 3] = 1

- 2 = y + 3.

- 5 = y.

The tangent line crosses the y - axis at (0, - 5).

The slope - intercept form (y = mx + b) of tangent line equation is y = x - 5.

The tangent line and normal line are perpendicular to each other.

The normal is just the same equation but with negative reciprocal of the slope.

The normal line is y = - x - 5.