find the values of `a` and `b` for the curve `x^2y+ay^2=b` if the point `(1,1)` is on its graph and the tangent line at `(1,1)` has the equation `9x+8y=17`
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First, determine the slope (m) of the tangent line. To do so, express the equation of tangent line in slope-intercept form.
Hence, the slope of the tangent line is `m=-9/8` .
Next, take the derivative of the curve to get dy/dx.
`d/(dx)(x^2y+ay^2) =d/dx (b)`
Now that we have the derivative of the curve, take note that the slope line tangent to the curve at the given point is equal to dy/dx.
`m = dy/dx`
Since the slope of the tangent line is -9/8 and is tangent to the curve at (1,1), plug-in these values to dy/dx.
`-9/8 = -(2*1*1)/(1^2+2a*1)`
Then, solve for a.
Next, solve for b. To do so, plug-in the point (1,1) and a=7/2 to the given curve.
Hence, `a= 7/2` and `b=9/2` and the equation of the curve is
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