Find the equation of the line tangent to the graph of x² - 3xy = 10 at point (1, -3)

Solution:

Given, the equation is x² - 3xy = 10.

We have to find the equation of the line tangent to the graph at point (1, -3).

The derivative of a function gives the slope of the tangent line, we can find ‘m’ in the equation of the line y = mx + b.

On differentiating,

d(x² - 3xy)/dx = d(10)/dx

By using product rule,

d(x² - 3xy)/dx = 2x - 3[y + xy’]

= 2x - 3y - 3x(dy/dx)

d(10)/dx = 0

So, 2x - 3y - 3x(dy/dx) = 0

2x - 3y = 3x(dy/dx)

dy/dx = (2x - 3y)/3x

At point (1, -3),

dy/dx = (2(1) - 3(-3))/3(1)

= (2 + 9)/3

= 11/3

dy/dx indicates the slope m.

So, m = 11/3.

The equation of the line in point-slope form is given by \((y-y_{1})=m(x-x_{1})\)

Here, m = 11/3, x₁ = 1, y₁ = -3

So, \((y-(-3))=\frac{11}{3}(x-1)\)

\((y+3)=\frac{11}{3}(x-1)\)

\(3(y+3)=11(x-1)\)

3y + 9 =11x - 11

11x - 3y = 9 + 11

11x - 3y = 20

11x - 3y - 20 = 0

Therefore, the equation of the line is 11x - 3y - 20 = 0.

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Find the equation of the line tangent to the graph of x² - 3xy = 10 at point (1, -3)

Summary:

The equation of the line tangent to the graph of x² - 3xy = 10 at point (1, -3) is 11x - 3y - 20 = 0.