0
If: y = x-3 and x2-3y2 = k
Then: x2-3(x-3)2 = k => -2x2+18x-27-k = 0
Using the discriminant of b2-4ac equals 0: 324-4*-2*(-27-k) = 0 gives k a value of 13.5
So when k = 13.5 there is only one point of contact
And when k > 13.5 there are no points of contact
But when k < 13.5 there are two points of contact
k = 0.1
They work out as: (-3, 1) and (2, -14)
It works out that line 3x-y = 5 makes contact with the curve 2x^2 +y^2 = 129 at (52/11, 101/11) and (-2, -11)
The line x-y = 2 intersects with the curve x^2 -4y^2 = 5 at (2.5, 1/3) and (3, 1) and by using the distance formula its length is 5/6
If: y = x^2 -10x +13 and y = x^2 -4x +7 Then: x^2 -10x +13 = x^2 -4x +7 Transposing terms: -6x +6 = 0 => -6x = -6 => x = 1 Substituting the value of x into the original equations point of contact is at: (1, 4)
(52/11, 101/11) and (-2, -11) Rearrange 3x-y = 5 into y = 3x-5 and substitute this into the curve equation and then use the quadratic equation formula to find the values of x which leads to finding the values of y by substituting the values of x into y = 3x-5.
-2
(2, -2)
It is (-0.3, 0.1)
k = 0.1
If: y = kx -2 and y = x^2 -8x+7 Then the values of k work out as -2 and -14 Note that the line makes contact with the curve in a positive direction or a negative direction depending on what value is used for k.
If the question is to do with a probability distribution curve, the answer is ONE - whatever the values of mu and sigma. The area under the curve of any probability distribution curve is 1.
If you mean the coordinates of the line x-y = 2 that intersects the curve of x2-4y2 = 5 Then the coordinates work out as: (3, 1) and (7/3, 1/3)
It works out that k must be greater than -11 or k must equal -11
They work out as: (-3, 1) and (2, -14)
It works out that line 3x-y = 5 makes contact with the curve 2x^2 +y^2 = 129 at (52/11, 101/11) and (-2, -11)
The line x-y = 2 intersects with the curve x^2 -4y^2 = 5 at (2.5, 1/3) and (3, 1) and by using the distance formula its length is 5/6