Find parameter p so that y - 3x is a tangent line to y - x^2 - p
You need to write the equation of the curve and the equation of the line such that: `y - x^2 - p = 0 and y - 3x = 0` .
Write the equation of the tangent line in the slope intercept form:
`y = 3x`
The slope of this line is m = 3 and it represents the derivative of the curve.
Differentiating the equation of the curve yields: `dy/dx = 2x` .
Since `dy/dx = m = 3 =gt 3 = 2x =gt x = 3/2` (the x coordinate of the tangency point)
Replacing x by `3/2` in the equation of the line yields:
`y = 3*3/2 = 9/2` (the y coordinate of the tangency point)
Replacing `x = 3/2` and y by `9/2` in the equation of the curve yields:
`9/2-9/4 - p = 0`
Adding p both sides yields: `p = 9/2-9/4 =gt p = 9/4`
Evaluating the parameter p yields: `p = 9/4` .