Question

12. - 11 points Y LARCALC9 2.5.031. Find the slope of the tangent line to the graph at the given point. (x2 + y2)2 = 4x?y, (1,1) Bifolium: (1,1) 10.5 -0.2 Need Help? Read It Watch it Talk to a Tutor

Answer 1

1. Given that (2+yj? = 4x+y; (1,1) (u? ty??czen?y=0 – 0 1 Delivative with respect to 'n' on both side in equation o - 4x^) . - du an olu dy 2 .0 a (n²t42) 1 du - a (laty? unty) = 0 low). Coul Avul => 1 (2+y137 al y] =0 21 2.++y+j'4 (n=440) _4( mal einen eigen 2) -- => 2 (1++% (21+zy dy) -4 (at dy ty Czu)) = 0 >> 4C*=+y) (x + y ) - 4(x+day+224) = 0 =4[(x+y+) (uty in Canada ant 2-24) =0 » (A++y?) (x+y a ten - Create + 2xy) = 0 >> 13+y?n tay x-dy - 2xy =0 om ay ( *y+ ry3 – 23) + x3 +y = x -244 = 0 ay (nry + y²_x²) = zay-x²_y²x:

zuy-x3-yu (ury + y² m²) . os me (201)61) –C?,200 1-0 + 13-12 2-1-1 ol- Tool) [m=o] Slope Tangent line equation is given by : (y-4)=m(*-*) (4-1) = 0 CX-1) => (4-1) =0 Ya! - Required tangent line is (y=1 2. Given that 3 Cu²+6²) ²100 (2²_42) , (-4,-2) 3(x² + y2) ²-100 (12_42) = 0 - 0 Delivative with respect to 'n' on both side in ea I du(3 (x²+ y2)2-100 (4²- y2)). du => 321_c2tyy?-100 _(*=y320 5) 3.2(2=tyy la Cestyy – 100 (zx-2y zo =) 6Cx?ty?) (antzy dy)-200 (haydy ) =0 on

- 200 + 192-48 +56.. 20 -96-24-100 Tangent like to given by (y-4.) = m (x-26) . (y=(->)) = (x-(-u)); 55 (972) = 14 (2+4). 554 +110 =144456 554 =14n+56-110 554 = 140-50 55 I (OR) . (orsly=0.25% -0.98 y = 1 / 5 (140-94) /