calculus
posted by mehak .
Find the coordinates of all points on the graph of y=23−x2 at which the tangent line passes through the point (10,26).please show the work i greatly appreciate it :)

y = 23  x^{2}
y' = 2x
So, we want all points where the line through (x,y) and (10,26) has slope 2x
(y26)/(x10) = 2x
(23x^{2}  26)/(x10) = 2x
3  x^{2} = 2x(x10)
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.
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