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find the equation of the tangent line to 9x^2+16y^2=52 through (2,-1) A. -9x+8y-26=0 B. 9x-8y-26=0 C. 9x-8y-106=0 D. 8x+9y-17=0 E. 9x+16y-2=0 the slope is 9/8 idk where to go from there...
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you got already the slope, and have a point on that tangent line so you're all set so y+1=9(x-2)/8 8y+8=9x-18 9x-8y-26=0 <---B
thanks
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