Posted by mehak on Sunday, October 30, 2011 at 9:38pm.
y = 23 - x^{2}
y' = -2x
So, we want all points where the line through (x,y) and (10,26) has slope -2x
(y-26)/(x-10) = -2x
(23-x^{2} - 26)/(x-10) = -2x
-3 - x^{2} = -2x(x-10)
-3 - x^{2} = -2x^{2} + 20x
x^{2} - 20x - 3 = 0
x = -0.149 or 20.149
At those values for x, we have
(x,y) = (-0.149,22.978) and (20.149,-382.982)
y' = 0.290 and -40.298
the line through (-0.149,22.978) and (10,26) has slope 3.022/10.149 = 0.29
the line through (20.149,-382.982) and (10,26) has slope 408.982/-10.149 = 40.298
Look slike those are our lines.