The gradient to the curve y = x2 - 8x + 7 is dy/dx = 2x - 8
The gradient of the tangent to the curve is, therefore, 2x - 8.
The gradient of the given line is k
Therefore k = 2x - 8. That is, k can have ANY value whatsoever.
Another Answer:-
If: y = kx-2 and y = x2-8x+7
Then: x2-8x+7 = kx-2 => x2-8x-kx+9 = 0
Use the discriminant of: b2-4ac = 0
So: (-8-k)2-4*1*9 = 0
Which is: (-8-k)(-8-k)-36 = 0 => k2+16k+28 = 0
Using the quadratic equation formula: k = -2 or k = -14 which are the possible values of k for the straight line to be tangent with the curve
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.
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
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)
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
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
2
Cotangent 32 equals tangent 0.031