There are 2 tangent lines to the parabola y = x^2 that pass through the point (0, -4). Find the coordinates of the points where these tangent lines intersect..

Question

There are two tangent lines to the parabola y = x² that pass through the point (0, -4). Find the coordinates of the points where these tangent lines intersect the parabola.

 

I know the equations for both tangent lines are -2x - 4 and 2x - 4

However, when I set x² = -2x - 4 and x² = 2x - 4, I can't seem to isolate x. What am I doing wrong?

 

p183, #59

ANSWER: (±2, 4)

Answer 1

A point on the parabola has coordinates (x, x²)

 

Slope of a tangent line through (x, x²) and (0, -4) is (x² - -4)/(x - 0)

 

                                                         = (x²+4)/x

 

But, slope of tangent line to the curve at (x, x²) is the derivative of     y = x², which is 2x.

 

So, (x²+4)/x = 2x

 

Therefore, x²+4 = 2x²

 

                   x² = 4

 

                   x = ±2

 

The points that we are looking for are (2, 2²) and (-2, (-2)²) also known as (2, 4) and (-2, 4).  

 

 

                                                        

Answer 2

The problem is symmetric about the y axis, so one of the lines will have positive slope with point of tangency on the right side of the parabola, while the other will have negative slope with point of tangency on the left.

 

The equation of the first line will be of the form  y =  m x  - 4  for some positive m.   Cal the x coordinate of the point of tangency x_t  .     The y coordinate of the point of tangency must be  m x_t - 4  (because it is on the line) but also  x_t ² (because it lies on the parabola).    This leads to the equation  m x_t -4 =  x_t ² .  

 

 The slope of the line and of the parabola must agree at x_t .   Since the slope of parabola is 2 x,    the equation 

  2 x_t = m can be deduced.     When this value of m is substituted into the first equation the resulting equation is

  2 x_t ² - 4 =  x_t ²   .   The solution is  x_t =  ± 2.      The corresponding value of m is m = ± 4  ,

   Plugging into  y =  x²  , the y value of the point of tangency is thus   ( ± 2)² = 4.

 

So the points of tangency are   (2, 4) and (-2, 4)

 

Answer 3

Mia,

 

Not sure where you got y = ±2x - 4.  Try y = ±4x - 4