+128408 We'll do (a) last
(b)
y^2 = x + 1 (1)
and
y = m (x + 3) square both sides
y^2 = m^2 (x + 3)^2 = m^2 (x^2 + 6x + 9) = m^2x^2 + 6m^2x + 9m^2 (2)
Set the y^2 ' s equal so that (1) = (2)
x + 1 = m^2x^2 + 6m^2x + 9m^2
0 = m^2x^2 + 6m^2x + 9m^2 - x - 1
0 = m^2x^2 + (6m^2 - 1)x + (9m^2 - 1)
(c)
It can be shown [ using implicit differentiation ] that the slope of any tangent line to the parabola is = 1 / [2y] = m
So
y = (m) (x + 3)
y =(1/ [2y]) ( x + 3)
2y^2 = x + 3 (3)
But
y^2 = x + 1 so
2y^2 = 2(x + 1) (4)
Set (3) equal to (4) and solve for x to find the x coordinate of the intersections
x + 3 = 2(x + 1)
x + 3 = 2x + 2
1 = x [this is the x coordinate for the intersection of the line and the parabola]
So....using this....we can find the y coordinates for the intersections
y^2 = x + 1
y^2 = 1 + 1
y^2 = 2
y = sqrt(2) or -sqrt(2)
[ these are the y coordinates for the intersection of the line and the parabola ]
So....m = 1/ [2y] = 1/ [2 *±sqrt (2) ] = 1 / [±sqrt(8)]
So...m^2 = [ 1 / ±sqrt (8) ] ^2 = 1/8
(d) The equations for the tangent lines are
y = [1 / sqrt(8) ] (x + 3) and
y = -(1/sqrt(8)] (x + 3)
And the intersection points are [ 1, sqrt(2) ] and [ 1, - sqrt(2) ]
(a) Here's the graph :
https://www.desmos.com/calculator/jaa8mpc6d7
