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If: kx+y = 4 and y = x^2 +8

Then: x^2 +8 = 4-kx or x^2 +8 -4+kx = 0 => x^2+4+kx = 0

The discriminant of the above quadratic equation must equal 0

So: k^2 -4*(4*1) = 0 => k^2-16 = 0

Therefore: k^2 = 16 and so the values of k are -4 and +4

Q: What are the values of k when kx plus y equals 4 is a tangent to the curve of y equals x squared plus 8 on the Cartesian plane showing work?

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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

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.

They are +/- 5*sqrt(2)

Points of intersection work out as: (3, 4) and (-1, -2)

Related questions

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

It is the Cartesian equation of an ellipse.

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, -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.

They are +/- 5*sqrt(2)

Points of intersection work out as: (3, 4) and (-1, -2)

-2

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

First find the slope of the circle's radius as follows:- Equation of circle: x^2 +10x +y^2 -2y -39 = 0 Completing the squares: (x+5)^2 + (y-1)^2 -25 -1 -39 = 0 So: (x+5)^2 +(y-1)^2 = 65 Centre of circle: (-5, 1) and point of contact (3, 2) Slope of radius: (1-2)/(-5-3) = 1/8 which is perpendicular to the tangent line Slope of tangent line: -8 Tangent equation: y-2 = -8(x-3) => y = -8x+26 Tangent equation in its general form: 8x+y-26 = 0