+9 2 Problems I got:
#38: for the graph of the function "y=1/x" In the first quarter a tangent formed through (a,1/a)
Express using "a"
1) the slope of the tangent
2) A. the equation of the tangent
B. Express using "a" the unions with the coordinate systems (x=0),(y=0)
C. show that the Surface that is created by the tangent when (x=0),(y=0) is not dependent on "a" and find it
D. Find (a,1/a) if the sum of the length of "x" and "y" when the tangent forms unions with the coordinate systems is 5
#39: for the graph of the funtion y=1/x^2 a tangent formed through (a,1/a^2)
A. Express using "a" the equation of the tangent
B. Express using "a" the unions with the coordinate systems (x=0),(y=0)
C. find "a" for whom the tangent creates "Isosceles triangle" with the coordinate systems
The problems from above are translated from my native language so I'm sorry if something can't be understood (I'll apretiate any tips on how I can write these problems in a better understoodable manner, or more efficently)
Thanks for anyone who helps!
+128577 #39: for the graph of the funtion y=1/x^2 a tangent formed through (a,1/a^2)
A. Express using "a" the equation of the tangent
B. Express using "a" the unions with the coordinate systems (x=0),(y=0)
C. find "a" for whom the tangent creates "Isosceles triangle" with the coordinate systems
A. dy/dx (1/x^2) = -2/x^3
Equation of the tangent at "a"
y - 1/ a^2 = (-2/a^3)(x - a)
y - 1/a^2 = -2x/a^3 + 2/a^2
y = -2x/a^3 + 3/a^2
y = [3a - 2x] / a^3
B. y intercept → x = 0 → y = 3/a^2 → (0, 3/a^2 )
x intercept → y = 0 → x = 3a/2 → ( 3a/2 , 0 )
C. Notice that many values of 'a' will produce isosceles triangles.......for instance......it we specify that the length of both tangent lines between the x and y axis = 5, we have
9/a^4 + 9a^2/4 = 25
9 + (9/4)a^6 = 25a^4
(9/4)a^6 - 25a^4 + 9 = 0
a≈ .78575 and a ≈ -.78575
Look at the graph, here......https://www.desmos.com/calculator/0twc2fqneu
Notice that two sides of the triangle formed by the points of the intersection of the tangent lines with both the x and y axis have lengths ≈ 5
+128577 38
1) dy / dx (1 / x) = -1/x^2
2) the slope of the tangent at "a" = -1/a^2
3) Equation of the tangent =
y - 1/a = ( -1 /a^2) ( x - a)
y = -x / a^2 + 2/a
y = [ 2a - x ] / a^2
B. I think you want to know where the tangent line intersects the y and x axis.
Wnen x = 0, y = 2/a this is the y intercept = (0, 2/a)
When y = 0, x = 2a this is the x intercept = (2a, 0 )
C. I think you want to find the area of a triangle bounded by the tangent line and both axis....if so, we have......
A = (1/2)(b)(h) = (1/2)(2a)(2/a) = 4 / 2 = 2 sq units........this is a constant area that does NOT depend upon "a"
D. We have
sqrt ( [2/a]^2 + [2a]^2 ) = 5 square both sides
4/a^2 + 4a^2 = 25 multiply through by a^2
4 + 4a^4 = 25a^2 rearrange
4a^4 - 25a^2 + 4 = 0
In the first quadrant, we have two solutions ... a ≈ .40536 and a ≈ 2.4669
So we have ( .40536, 1/.40536) and (2.4669, and 1/ 2.4669 )
Here's a graph of both tangent line solutions......note that the total length of both tangent lines between their intersections with the x and y axis ≈ 5
https://www.desmos.com/calculator/km8gajbnnl
+128577 #39: for the graph of the funtion y=1/x^2 a tangent formed through (a,1/a^2)
A. Express using "a" the equation of the tangent
B. Express using "a" the unions with the coordinate systems (x=0),(y=0)
C. find "a" for whom the tangent creates "Isosceles triangle" with the coordinate systems
A. dy/dx (1/x^2) = -2/x^3
Equation of the tangent at "a"
y - 1/ a^2 = (-2/a^3)(x - a)
y - 1/a^2 = -2x/a^3 + 2/a^2
y = -2x/a^3 + 3/a^2
y = [3a - 2x] / a^3
B. y intercept → x = 0 → y = 3/a^2 → (0, 3/a^2 )
x intercept → y = 0 → x = 3a/2 → ( 3a/2 , 0 )
C. Notice that many values of 'a' will produce isosceles triangles.......for instance......it we specify that the length of both tangent lines between the x and y axis = 5, we have
9/a^4 + 9a^2/4 = 25
9 + (9/4)a^6 = 25a^4
(9/4)a^6 - 25a^4 + 9 = 0
a≈ .78575 and a ≈ -.78575
Look at the graph, here......https://www.desmos.com/calculator/0twc2fqneu
Notice that two sides of the triangle formed by the points of the intersection of the tangent lines with both the x and y axis have lengths ≈ 5
+128577 39C.....additional commment
If we are only talking about a triangle in the first quadrant bounded by the tangent line and the two axis.....an isosceles triangle can be formed when :
3/a^2 = 3a / 2 cross-multiply
6 = 3a^3
2 = a^3
a = (2)^(1/3)
See the graph, here : https://www.desmos.com/calculator/rmr2qlx29k
Two sides of the triangle wil lbe about 1.89 units in length......these sides will fall on both of the coordinate axis
