To find the equation of the level curve, we need to set the function f(x,y) equal to a constant value, say k. This will give us a curve that represents all the points (x,y) where the function has the same value, or in other words, the level curve.
In this case, we have f(x,y) = yx^2 - y^2. Setting this equal to a constant k, we get yx^2 - y^2 = k. We can rearrange this equation to get y(x^2 - 1) = k, and then solve for y to get y = k/(x^2 - 1).
Now, we can substitute the given point (2,1) in this equation to find the specific value of k. Plugging in x=2 and y=1, we get 1 = k/(4-1), which gives us k = 3. So the equation of the level curve is y = 3/(x^2 - 1).
To find the equation of the tangent line to this level curve at the point (2,1), we can use the derivative of the function f(x,y) with respect to x. This will give us the slope of the tangent line at any given point on the curve.
The derivative of f(x,y) with respect to x is 2yx, so at the point (2,1), the slope of the tangent line is 2(1)(2) = 4. We can use this slope and the given point (2,1) to find the equation of the tangent line using the point-slope form: y - y1 = m(x - x1). Plugging in the values, we get y - 1 = 4(x - 2), which simplifies to y = 4x - 7.
So the equation of the tangent line to the level curve at the point (2,1) is y = 4x - 7. Your answer of 10 = 4x + 2y is incorrect. I hope this helps clarify the process for finding both the equation of the level curve and the tangent line.
