The paraboloid z = 6 − x − x² − 7y² intersects the plane x = 1 in a parabola. Find parametric equations in terms of t for the tangent line to this parabola at the point (1, 2, −24). (Enter your answer as a comma-separated list of equations. Let x, y, and z be in terms of t.)

Answer:First you need to know is the equation of the parabola in order to get the equation of the tangent line. So we replace x = 1 in the paraboloid equation and we get:[tex]z = 6 -1 -1 -7y^{2} \\z = 4 - 7y^{2}[/tex]So now that we have the parabola's equation, we calculate the slope of the tangent line deriving and replacing with the point (2,-24) (this point doesn't have the x term because we already used it and we are in terms of y and z).[tex]z' = -14y\\slope = m = -14*2 = -28[/tex]Now we have the next equation:[tex]z = -28y + b[/tex]In order to calculate the term 'b', we replace (y,z) with the point (2,-24):[tex]-24 = -28*2 + b\\b = 32[/tex]Then, we finally get the tangent line equation as follow:[tex]z = -28y+32[/tex]Finally, in order to convert the variables in terms of t, we just replace 't' in any variable. In this case I will replace in y because is convenient.y = t,z = -28t+32,x = 1 (because is always a constant so It doesn't depend of any variable)