Question
Asked Oct 8, 2019

48(x^2+y^2)^2=625xy^2;(3,4)

Write an equation for the tangent line at the point (3,4)

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Expert Answer

Step 1

Given,

48(x2y2625xy2 and point (3,4)
=
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48(x2y2625xy2 and point (3,4) =

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Step 2

Now, slope of tangent line is given by dy/dx at  point (3,4).

48(x222625xy2
Differentiating on both sides with respect to x, we get
d(48(x2y2)2d(625xy2)
dx
dx
d((x2y22)
.d(xy2)
625
48
dx
dx
48 x 2(x2y2d(x2+y2)
dx
d(y2)
y2
dx
, d (x)
625 x
11
dx
dy
96(x2y22x + 2y
dx
dy
625 2xy
dy
су
dy
192y2x192y3
dx
dy
192x3192x2y-
1250xy 625y
dx
dy
dy
dy
625y2 - 192x3 - 192y2x
dx
192x2y
1250xy
dx
192y3
dx
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48(x222625xy2 Differentiating on both sides with respect to x, we get d(48(x2y2)2d(625xy2) dx dx d((x2y22) .d(xy2) 625 48 dx dx 48 x 2(x2y2d(x2+y2) dx d(y2) y2 dx , d (x) 625 x 11 dx dy 96(x2y22x + 2y dx dy 625 2xy dy су dy 192y2x192y3 dx dy 192x3192x2y- 1250xy 625y dx dy dy dy 625y2 - 192x3 - 192y2x dx 192x2y 1250xy dx 192y3 dx

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Step 3

Further calcu...

625y2 192x3 -192y2x
dy
192x2y 192y3 1250xy
dx
So, slope dy
dx\(3,4)
625(4)2-192(3)3-192(4)2 (3)
192(3)2(4)+192(4)3-1250(3) (4)
22
11
21
So, equation of the tangent line passes through the point (3,4) is
22
(x 3)
21
у — 4 3
21у - 84 3 —22х + 66
150
21y22x
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625y2 192x3 -192y2x dy 192x2y 192y3 1250xy dx So, slope dy dx\(3,4) 625(4)2-192(3)3-192(4)2 (3) 192(3)2(4)+192(4)3-1250(3) (4) 22 11 21 So, equation of the tangent line passes through the point (3,4) is 22 (x 3) 21 у — 4 3 21у - 84 3 —22х + 66 150 21y22x

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Tagged in

Math

Calculus

Derivative