normal and tangent

mr.burger

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Jun 12, 2005
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Find the area of the triangle formed by the x-axis and the normal and tangent lines to the graph of x^(2)+y^(2)=25 at the point (3,4).
 

Gene

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Oct 8, 2003
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The tangent line is y = (-3/4)(x-3)+4
The x intercept is at
-3/4*x+9/4+4 = 0
3x=25
x=25/3
Assuming the normal is to the x axis the triangle is P(3,4), P(3,0) and P(25/3,0)
The area is .5*4*(25/3-3) = 32/3
That will give you a check if you feel it should be an integral.
 

mr.burger

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Jun 12, 2005
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i still dont understand is the circle around the point (3,4)? if it is then i dont understand how the x-axis the normal and tangent of the circle make a triangle because with the points you gave me i can understand it doesnt fit not having a graphing calc doesnt help either
 

Gene

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The equation says the circle is around the origin and has a radius of 5. (note 3²+4²=5²) Point (3,4) is on the circle. A line from the origin has a slope of 4/3. The tangent is perpendicilar to that line so it has a slope of the negative reciprical, (-3/4) and goes thru P(3,4) so it's equation is
y-4 = (-3/4)(x-3)
The x intercept of the tangent (y=0) is the second point of the triangle. If the normal is to the x axis then it starts at the third point P(3,0) and goes up to P(3,4)
We are in the first quadrent so sketching a quarter of a circle and marking those three points shouldn't be too hard. If you don't have graph paper you can print some from
http://www.mathematicshelpcentral.com/graph_paper.htm
 

tkhunny

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mr.burger said:
i still dont understand is the circle around the point (3,4)? if it is then i dont understand how the x-axis the normal and tangent of the circle make a triangle because with the points you gave me i can understand it doesnt fit not having a graphing calc doesnt help either
Here we have an example that you may need to back up, leave calculus alone for a while, and brush up on your analytic geometry. There are reasons why course listings include prerequisites.
 
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