Verify that the given point lies on the curve. Determine an equation of the line

Question

Verify that the given point lies on the curve. Determine an equation of the line tangent to the curve at the given point. 6x2 + 6xy + 4y2 = 96; (2, 3) The point (2, 3) is in the first quadrant. The point (2, 3) is in the domain of the implicit function. The value of 6x2 + 6xy + 4y2 is less than 96 when x is 2 and y is 3. Write the equation for the tangent line in slope-intercept form. Y = (Use integers or fractions for any numbers in the equation.)

Explanation / Answer

6x^2+6xy+4y^2 x = 2, y = 3 =>6x^2+6xy+4y^2 = 96 So given point lies on the curve Now differentiating 6x^2+6xy+4y^2=96 w.r.t x we get 12x + 6y + 6xdy/dx + 8ydy/dx = 0 =>dy/dx = -(12x+6y)/(6x+8y) = m x = 2, y = 3 => m = -42/36 = -7/6 Equation of tangent is, y-3 = (-7/6)(x-2) y = (-7/6)x + 16/3

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