"Lemniscate" just refers to the type of shape formed by the graph of this equation. You'll probably run into this term again when you're doing polar coordinates. Don't worry about that part now.
The tangent line is the line through a point on the curve which has a slope equal to the derivative at that point. What sort of slope does an horizontal line have? You need to find the points on the curve where dy/dx has that value, so the slope at that point has that value, so the tangent line through that point is horizontal.
To find where dy/dx equals zero, differentiate implicitly, and solve for dy/dx.
. . . . .8(x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup> = 169(x<sup>2</sup> - y<sup>2</sup>)
. . . . .16(x<sup>2</sup> + y<sup>2</sup>)(2x + 2y[dy/dx]) = 169(2x - 2y[dy/dx])
Simplify by dividing through by 2 to get:
. . . . .(16x<sup>2</sup> + 16y<sup>2</sup>)(x + y[dy/dx]) = 169x - 169y[dy/dx]
. . . . .16x<sup>3</sup> + 16xy<sup>2</sup> + 16x<sup>2</sup>y[dy/dx] + 16y<sup>3</sup>[dy/dx] = 169x - 169y[dy/dx]
. . . . .169y[dy/dx] + 16x<sup>2</sup>y[dy/dx] + 16y<sup>3</sup>[dy/dx] = 169x - 16x<sup>3</sup> - 16xy<sup>2</sup>
Complete the computations to solve for "dy/dx =". Set the result equal to zero, and solve, recalling that a fraction is zero when the numerator is zero.
The result (as far as I took it, anyway) is fairly messy, and you'll want to check my work, but I hope the above helps a bit.
Eliz.
