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

So, you have to take the derivative first of 9x^2+16y^2=52.
 
18x + (32y)(y') = 0 
32y (y') = -18x 
y' = -18x / 32y 

plug in for 0, to find the slope which will get you - 9/16. 
and you know that the tangent of the line formula is y - y1 = m (x -x1), and the problem gave you the points for y1 and x1. 

so the line tangent is y = - 9/16 (x - 2) - 1

I hope my answer has come to your help. Thank you for posting your question here in .

I’m pretty sure the slope would be 1 since the x and y values increase by 1