Find the equation of the tangent and normal lines to the given curves at the indicated point?

Question

1. x^2+y^2=25; (-3,-4)

2. x^2-6x+2y-8=0; x=3

Answer 1

(1)

Step 1:

The equation is and the point is .

Consider .

Differentiate on each side with respect to .

Slope of the tangent is derivative of the function at the given point.

.

Slope of the tangent line is .

Point slope form of line equation is .

Substitute and in the above equation.

Tangent line is  .

Step 2:

Normal line is perpendicular to tangent line then

slope of tangent line slope of normal line is equal to .

Let slope of the normal as .

Therefore,

Point slope form of line equation is .

Substitute and in the above equation.

Normal line equation is .

Solution:

Tangent line is .

Normal line equation is .

Answer 2

(2)

Step 1:

The equation is .

Apply derivative on each side with respect to .

.

Slope of the function is derivative of the function at that point.

Slope of the tangent line at .

.

Slope of the tangent line is .

Step 2:

Substitute in .

.

The tangent at the point .

Point slope form of line equation is .

Substitute and in the above equation.

Tangent line equation is .

Step 3:

The normal line is perpendicular to the tangent line.

The equation of normal line at is , Where is slope of the tangent line.

Substitute and in the above equation.

Normal line equation is .

Solution:

Tangent line equation is .

Normal line equation is .