Use implicit differentiation to find an equation of the tangent line to the eclipse defined by 2x^2+3xy+2y^2=46 at point (1,4). Solve it step by step

Given equation of the ellipse:
2x^2 + 3xy + 2y^2 = 46

Step 1: Differentiate both sides of the equation implicitly with respect to x :
\frac{d}{dx} (2x^2) + \frac{d}{dx} (3xy) + \frac{d}{dx} (2y^2) = \frac{d}{dx} (46)
4x + 3y + 3x \frac{dy}{dx} + 4y \frac{dy}{dx} = 0
3x \frac{dy}{dx} + 4y \frac{dy}{dx} = -4x - 3y
\frac{dy}{dx}(3x + 4y) = -4x - 3y
\frac{dy}{dx} = \frac{-4x - 3y}{3x + 4y}

Step 2: Find the slope of the tangent line at point (1,4) by substituting x = 1 and y = 4 into the equation:
\frac{dy}{dx} \bigg|_{(1,4)} = \frac{-4(1) - 3(4)}{3(1) + 4(4)}
\frac{dy}{dx} \bigg|_{(1,4)} = \frac{-4 - 12}{3 + 16}
\frac{dy}{dx} \bigg|_{(1,4)} = \frac{-16}{19}

Step 3: The slope of the tangent line at point (1,4) is m = -\frac{16}{19} .
Since the tangent line passes through the point (1,4), we can write the equation of the tangent line in point-slope form:
y - y_1 = m(x - x_1)
y - 4 = -\frac{16}{19}(x - 1)

Step 4: Simplify the equation and write the equation of the tangent line in slope-intercept form:
y - 4 = -\frac{16}{19}x + \frac{16}{19}
y = -\frac{16}{19}x + \frac{16}{19} + 4
y=-\frac{16}{19}x+\frac{92}{19}
or by rearranging terms:

19y=-16x+92\xrightarrow{}16x+19y=92 .

{Answer:} The equation of the tangent line to the ellipse at point (1,4) is :
y=-\frac{16}{19}x+\frac{92}{19}
or
16x+19y = 92