# 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`

*print*Print*list*Cite

First, determine the slope (m) of the tangent line. To do so, express the equation of tangent line in slope-intercept form.

`9x+8y=17`

`9x-9x+8y=-9x+17`

`8y=-9x+17`

`(8y)/8=(-9x+17)/8`

`y=-9/8x+17/8`

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)`

`x^2*dy/dx+y*2x+a*2y(dy)/(dx)=0`

`x^2(dy)/(dx)+2xy+2ay(dy)/(dx)=0`

`x^2(dy)/(dx)+2ay(dy)/(dx)=-2xy`

`(dy)/(dx)(x^2+2ay)=-2xy`

`(dy)/(dx)=-(2xy)/(x^2+2ay)`

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)`

`-9/8=-2/(1+2a)`

Then, solve for a.

`-9(1+2a)=-2*8`

`-9-2a=-16`

`-2a=-16+9`

`-2a=-7`

`a=7/2`

Next, solve for b. To do so, plug-in the point (1,1) and a=7/2 to the given curve.

`x^2y+ay^2=b`

`1^2*1+7/2*1^2=b`

`1+7/2=b`

`2/2+7/2=b`

`9/2=b`

**Hence, `a= 7/2` and `b=9/2` and the equation of the curve is **

**`x^2y+7/2y^2=9/2` .**