If you mean: y = kx-2 and y = x^2-8x+7
Then: x^2-8x+7 = kx-2
=> x^2-(8x-kx)+7+2 = 0
=> x^2-8x+kx+9 = 0
For the tangent to touch the curve the discriminant of b^-4ab must = 0
So: (8+k)^2-4*(9*1) = 0 => (8+k)^2 -36 = 0 => (8+k)^2 = 36
Square root both sides and then subtract 8 from both sides:
k = - or + 6 -8
Therefore possible values of k are: k = -2 or k = -14
They are +/- 5*sqrt(2)
k = 0.1
In trig, the secant squared divided by the tangent equals the hypotenuse squared divided by the product of the opposite and adjacent sides of the triangle.Details: secant = hypotenuse/adjacent (H/A) and tangent = opposite/adjacent (A/O);Then secant2/tangent = (H2/A2)/(O/A) = H2/A2 x A/O = H2/AO.
Circle equation: x^2 +y^2 -8x +4y = 30 Tangent line equation: y = x+4 Centre of circle: (4, -2) Slope of radius: -1 Radius equation: y--2 = -1(x-4) => y = -x+2 Note that this proves that tangent of a circle is always at right angles to its radius
If the line y = 2x+1.25 is a tangent to the curve y^2 = 10x then it works out that when x = 5/8 then y = 5/2
They are +/- 5*sqrt(2)
(2, -2)
k = 0.1
In trig, the secant squared divided by the tangent equals the hypotenuse squared divided by the product of the opposite and adjacent sides of the triangle.Details: secant = hypotenuse/adjacent (H/A) and tangent = opposite/adjacent (A/O);Then secant2/tangent = (H2/A2)/(O/A) = H2/A2 x A/O = H2/AO.
-2
This is not possible, since the point (4,6) lies inside the circle : X2 + Y2 = 16 Tangents to a circle or ellipse never pass through the circle
It is (-0.3, 0.1)
If the line y = 2x+1.25 is a tangent to the curve y^2 = 10x then it works out that when x = 5/8 then y = 5/2
Circle equation: x^2 +y^2 -8x +4y = 30 Tangent line equation: y = x+4 Centre of circle: (4, -2) Slope of radius: -1 Radius equation: y--2 = -1(x-4) => y = -x+2 Note that this proves that tangent of a circle is always at right angles to its radius
2
Equation of circle: x^2 +y^2 -8x -y +5 = 0Completing the squares: (x-4)^2 +(y-0.5)^2 = 11.25Centre of circle: (4, 0.5)Slope of radius: -1/2Slope of tangent: 2Equation of tangent: y-2 = 2(x-1) => y = 2xNote that the above proves the tangent of a circle is always at right angles to its radius
Cotangent 32 equals tangent 0.031