for x/y + 5x -7 =-3/4y find the equation of the tangent line given the equation above and the point P(1,2)?

(x/y) + 5x - 7 = - (3/4).y

x + 5xy - 7y = - 3y²/4

4x + 20xy - 28y = - 3y²

3y² - 28y + 20xy + 4x = 0

3y² - 28y + 20xy + 4x = 0 → when x is constant: → (6y - 28 + 20x).dy = 0

3y² - 28y + 20xy + 4x = 0 → when y is constant: → (20y + 4).dx = 0

(6y - 28 + 20x).dy + (20y + 4).dx = 0

(6y - 28 + 20x).dy = - (20y + 4).dx

(3y - 14 + 10x).dy = - (10y + 2).dx

dy/dx = - (10y + 2)/(3y - 14 + 10x) → given point P (1 ; 2)

dy/dx = - (20 + 2)/(6 - 14 + 10)

dy/dx = - 22/2

dy/dx = - 11 ← this is the slope of tha tangent line to the curve at x = 1

The typical equation of a line is: y = mx + y₀ → where m: slope and where y₀: y-intercept

The slope of the tangent line is (- 11), so the equation of the tangent line becomes: y = - 11x + y₀

The tangent line passes through P (1 ; 2), so these coordinates must verify the equation of the tangent line.

y = - 11x + y₀

y₀ = y + 11x → you substitute x and y by the coordinates of the point P (1 ; 2)

y₀ = 2 + 11

y₀ = 13

→ The equation of the tangent line at P is: y = - 11x + 13

Begin by writing the equation in general form.

x/y + 5x - 7 = -3/4y

4x + 20xy - 28y = -3y² ... (y ≠ 0)

3y² + 20xy + 4x - 28y = 0

Point P does satisfy the equation, which makes our job easier. This is a second-degree equation in x and y. The tangent line must include P, and it would be a bit simpler here to substitute for x, so let this be the line of tangency:

x = k(y - 2) + 1

x = ky + 1 - 2k

3y² + 20xy + 4x - 28y = 0

3y² + 20y(ky + 1 - 2k) + 4(ky + 1 - 2k) - 28y = 0

(3 + 20k)y² + (-8 - 36k)y + (4 - 8k) = 0

It is a quadratic equation in y. Let the discriminant equal zero.

(-8 - 36k)² - 4(3 + 20k)(4 - 8k) = 0

(2 + 9k)² - (3 + 20k)(1 - 2k) = 0

121k² + 22k + 1 = 0

(11k + 1)² = 0

k = -1/11

Substitute that into the tangent line equation.

x = ky + 1 - 2k

x = (-1/11)y + 1 - 2(-1/11)

11x + y - 13 = 0

MAny ways here:

Performing implicit differentiation, find y'. this is the slope m.

At P, this slope is thus ...? Thus Y = mX + B

From P, you find B.

Done!

Another way, view itr all as x = x(y) and find x'(y). No implicit differentiation required...

If need be, show your steps and answers here. Then we can take it from there.