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 December 11th, 2010, 06:25 AM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 verify that the given point is on the curve and find the lin verify that the given point is on the curve and find the lines that are ( a ) tangent (b) normal to curve at the given point ? x^2 + xy - y^2 = 1 point ( 2,3) >>>> a ) 2x +xy'+y-2yy' = 0 y' = -2x - y / x - y m = y' = -2(2) - 3 / 2-2(3) = 1.75 the Eqation of tangent .. is : y-3 = 1.75(x-2 ) about (b) how i solve it ?
 December 11th, 2010, 07:41 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: verify that the given point is on the curve and find the You are given the curve $x^2+xy-y^2=1$ and the point (2,3). To verify that the point is on the curve, make sure $2^2+2\cdot3-3^2=1$ To find the line tangent to the curve at the given point, find y': $2x+xy'+y-2yy'=0$ $y'(x-2y)=-(2x+y)$ $y'=\frac{2x+y}{2y-x}$ $y'(2,3)=\frac{2\cdot2+3}{2\cdot3-2}=\frac{7}{4}$ Somehow you got the correct slope, but your formula for y' was slightly wrong. With the slope and the point, we have: $y-3=\frac{7}{4}(x-2)$ I recommend putting the line into y = mx + b form: $y=\frac{7}{4}x-\frac{1}{2}$ To find the normal line, use the fact that when two slopes are perpendicular, what is their product?

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