A line has equation 4x + y + k = 0 and a curve has equation y = kx? + 3, where k is a constant. a) Find the two values of k for which the line is a tangent to the curve b) Find the equation of the straight line joining the points on the curve for these two values of k.A line has equation 4x + y + k = 0 and a curve has equation y = kx? + 3, where k is a constant. a) Find the two values of k for which the line is a tangent to the curve b) Find the equation of the straight line joining the points on the curve for these two values of k.

It seems like there was a mistake in your question as it was repeated twice. However, I'll answer the question as follows:

a) For the line to be a tangent to the curve, the discriminant of the quadratic formed when the equations are equated should be zero (since a tangent line touches the curve at exactly one point).

First, let's rearrange the equation of the line to y = -4x - k.

Then, equate this to the equation of the curve: -4x - k = kx^2 + 3.

Rearrange this to form a quadratic equation: kx^2 + 4x + (k + 3) = 0.

The discriminant of a quadratic ax^2 + bx + c is given by b^2 - 4ac. For the line to be a tangent, the discriminant should be zero. So, we have:

(4)^2 - 4k(k + 3) = 0.

Solving this quadratic equation for k gives k = -3 or k = -4/3.

b) The points of tangency are the solutions to the equation of the curve for these values of k. Substituting k = -3 into the curve equation gives y = -3x^2 + 3, and solving for x gives x = ±1. So the points of tangency are (-1, 6) and (1, 6).

Substituting k = -4/3 into the curve equation gives y = -(4/3)x^2 + 3, and solving for x gives x = ±sqrt(3). So the points of tangency are (-sqrt(3), 5) and (sqrt(3), 5).

The equation of the line joining these points can be found using the formula for the slope of a line (y2 - y1) / (x2 - x1) and the point-slope form of a line equation y - y1 = m(x - x1). However, since the y-coordinates of the points are the same, the line is horizontal and its equation is simply y = 6.