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

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

2. x^2-6x+2y-8=0; x=3
asked Aug 6, 2015 in PRECALCULUS by anonymous
reshown Aug 6, 2015 by bradely

2 Answers

0 votes

(1)

Step 1:

The equation is image and the point is image.

Consider image.

Differentiate on each side with respect to .

image

image

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

image

image.

Slope of the tangent line is image.

Point slope form of line equation is image.

Substitute image and image in the above equation.

image

image

Tangent line is image .

Step 2:

Normal line is perpendicular to tangent line then

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

Let slope of the normal as image.

Therefore,

image

Point slope form of line equation is image.

Substitute image and image in the above equation.

image

image

Normal line equation is image.

Solution:

Tangent line is image .

Normal line equation is image.

answered Aug 6, 2015 by cameron Mentor
edited Aug 6, 2015 by cameron
0 votes

(2)

Step 1:

The equation is image.

image

Apply derivative on each side with respect to .

image

image

image

image.

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

Slope of the tangent line at image.

image

image.

Slope of the tangent line is image.

Step 2:

Substitute image in image.

image

image

image

image

image

image.

The tangent at the point image.

Point slope form of line equation is image.

Substitute image and image in the above equation.

image

Tangent line equation is image.

Step 3:

The normal line is perpendicular to the tangent line.

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

Substitute image and image in the above equation.

image

Normal line equation is image.

Solution:

Tangent line equation is image.

Normal line equation is image.

answered Aug 6, 2015 by Sammi Mentor

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