2x+2yy′=02x+2yy'=02x+2yy′=0 hence y′=−2x2y=−xyy'=-\frac{2x}{2y}=-\frac{x}{y}y′=−2y2x=−yx then y′=5/4y'=5/4y′=5/4 if (−5,4)(-5,4)(−5,4) , hence equation of the tangent is y−4=54(x+5)y-4=\frac{5}{4}(x+5)y−4=45(x+5) or 5x−4y+41=05x-4y+41=05x−4y+41=0
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