# Equation of Tangent to the Line

• johnq2k7
In summary, the equation of the tangent line to a given line is typically written in the form y = mx + b, where m represents the slope of the tangent line and b represents the y-intercept of the line. The point of tangency on a line can be found by first taking the derivative of the line's equation, setting it equal to the slope of the tangent line, and solving for the x-value. A line can only have one tangent at any given point, and a line and a curve intersect at a tangent if their slopes are equal at that point. The equation of a tangent line can never be a vertical line because a tangent line must have a defined slope to touch a curve at a specific point.
johnq2k7
3.) Give the equation of the line tangent to the curve at the given point.

a.) (y)(tan^-1 x) = x*y at (sqrt(3),0)

b.) ln y = x^2 +(2)*e^x at (0, e^2)

Here is some of my work process:

i have no idea how to solve for y in a.)

but for b.) I took the expontential of both sides

then i got y= e^(x^2) + e^(2*e^x)

do I find the derivate of y to get the slope

but how do i plug in the values of the pt.

do i use the y- yl= m(x-x1) formula to find the tangent to the curve?

please help me with these problems, I'm clueless for part a, and i have a bit of an understanding on how to do part b.)

johnq2k7 said:
3.) Give the equation of the line tangent to the curve at the given point.

a.) (y)(tan^-1 x) = x*y at (sqrt(3),0)

b.) ln y = x^2 +(2)*e^x at (0, e^2)

Here is some of my work process:

i have no idea how to solve for y in a.)

but for b.) I took the expontential of both sides

then i got y= e^(x^2) + e^(2*e^x)

do I find the derivate of y to get the slope

but how do i plug in the values of the pt.
Put the x value of the point into your formula for the derivative!

do i use the y- yl= m(x-x1) formula to find the tangent to the curve?
Yes!

please help me with these problems, I'm clueless for part a, and i have a bit of an understanding on how to do part b.)
You may have miscopied part (a). If y is not 0, you can divide through by it and get tan-1(x)= x which means y is not a function of x. I think that there are two points, one positive and the other negative, for which tan-1(x)= x and so the equation is true for all y. Of course, it is true for all x if y= 0 and so its graph is those three lines. If that is really is what (a) says, then since $tan^{-1}(\sqrt{3})\ne \sqrt{3}$, the curve in a neighbor hood of the point $(\sqrt{3},0)$ is the line y= 0 and so its tangent is the line y= 0.

## 1. What is the equation of the tangent line to a given line?

The equation of the tangent line to a given line is typically written in the form y = mx + b, where m represents the slope of the tangent line and b represents the y-intercept of the line. The slope of the tangent line can be calculated by taking the derivative of the given line's equation at the point of tangency.

## 2. How do you find the point of tangency on a line?

The point of tangency on a line can be found by first taking the derivative of the line's equation. Then, set the derivative equal to the slope of the tangent line and solve for the x-value. Plug this x-value into the original equation to find the corresponding y-value. The resulting point (x, y) is the point of tangency on the line.

## 3. Can a line have multiple tangents?

No, a line can only have one tangent at any given point. This is because a tangent is defined as a line that touches a curve at only one point and has the same slope as the curve at that point.

## 4. How do you determine if a given line and curve intersect at a tangent?

A line and a curve intersect at a tangent if the line is tangent to the curve at a specific point. This can be determined by checking if the slope of the line at that point is equal to the slope of the curve at the same point.

## 5. Can the equation of a tangent line ever be a vertical line?

No, the equation of a tangent line can never be a vertical line. This is because a vertical line has an undefined slope, while a tangent line must have a defined slope to touch a curve at a specific point.

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