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How do a secant line and a tangent line that pass through a common point on a graph differ?

2007-03-29 04:53:24 · 6 answers · asked by Anonymous in Science & Mathematics Mathematics

6 answers

Well, in the neighborhood of this common point, the tangent line only intersects the graph in one point, while the secant intersects in two. I said "neighborhood" because you could have a tangent line that ends up intersecting the graph at another point down the road (think of a sine graph as an example).

2007-03-29 05:01:55 · answer #1 · answered by Kathleen K 7 · 0 0

A tangent line can touch a circle at only one point. A secant cuts a circle at two points. Thus the secant and tangent can share a common point only if the point is outside a circle or at the point of tangency.

2007-03-29 05:08:25 · answer #2 · answered by ironduke8159 7 · 0 0

Let the graph of the function be y = f(x)

For a tangent line dy/dx = slope of tangent line
For a secant line dy/dx != slope of secant line

2007-03-29 05:01:20 · answer #3 · answered by Nishit V 3 · 0 0

That common point is one of the vertices of a triangle.
One other vertex is the point of tangency, the remaining vertex is an origin, forming the angle on which the tan and sec are calculated.

Try drawing it out as described above.Use a circle with center at (0,0), P as point of tangency at 45 degrees. Where the tangent line crosses the X axis is the common point. The length from (0,0) to this point is the sec of theta, the angle formed by the radius to point P.

2007-03-29 05:06:29 · answer #4 · answered by davidosterberg1 6 · 0 0

a tangent only touches the curve at one point...
a secant line touches two or more points

if you think of a circle ....
the line that touches the circle at only one point, it is a tangent....
any line that intersects the circle at more than 1 point (2 points... a chord or diameter) is a secant

2007-03-29 05:05:20 · answer #5 · answered by Brian D 5 · 1 1

extremely tough thing look on from google and yahoo that will could actually help

2014-07-21 19:00:08 · answer #6 · answered by Anonymous · 0 0

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