What Is the Equation for a Tangent Line Parallel to a Given Line?

AI Thread Summary
To find the equation of a tangent line to the curve y = Sqrt(2x - 1) that is parallel to the line x - 3y = 16, first determine the slope of the given line, which is 1/3. Next, calculate the derivative of the curve to find the slope of the tangent line. Set the derivative equal to 1/3 to find the corresponding x-value on the curve. Use the point-slope form of the equation to derive the equation of the tangent line. Understanding these steps is crucial for solving similar problems in calculus.
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Homework Statement



What is the equation of the tangent line to the curve y = Sqrt( 2x - 1) where the tangent line is parallel to the line x - 3y = 16

The Attempt at a Solution



My teacher briefly mentioned to find the derivative of the given, and then plug it into the slope-y intercept formula? I got confused from there because he used y-y=m(x-x)
 
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DespicableMe said:

Homework Statement



What is the equation of the tangent line to the curve y = Sqrt( 2x - 1) where the tangent line is parallel to the line x - 3y = 16

The Attempt at a Solution



My teacher briefly mentioned to find the derivative of the given, and then plug it into the slope-y intercept formula? I got confused from there because he used y-y=m(x-x)
The slope-intercept form is often written as y = mx + b.

The point-slope form is written as y - y0 = m(x - x0). This form is useful if you know a point (x0, y0) on the line, and the slope m of the line.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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