Find the equation of the tangent line at the given point on the curve.
-$x y^{2}=4 ;(4,1)$
A) $y=-8 x+33$
B) $y=-\frac{1}{8} x+\frac{3}{2}$
C) $y=-\frac{1}{2} x+3$
D) $y=-2 x+9$

Question

Quizplus · Accepted Answer

To find the equation of the tangent line, we need to find the derivative of the curve at the given point. Using implicit differentiation, we get:$$\frac{d}{dx}(xy^2) = \frac{d}{dx}(4)$$$$y^2 + 2xy\frac{dy}{dx} = 0$$$$\frac{dy}{dx} = -\frac{y^2}{2xy} = -\frac{y}{2x}$$At the point (4,1), the slope of the tangent line is:$$\frac{dy}{dx} = -\frac{1}{2(4)} = -\frac{1}{8}$$So the equation of the tangent line is:$$y - 1 = -\frac{1}{8}(x - 4)$$$$y = -\frac{1}{8}x + \frac{3}{2}$$

Find the Equation of the Tangent Line at the Given xy2=4;(4,1)x y^{2}=4 ;(4,1)xy2=4;(4,1)

Find the Equation of the Tangent Line at the Given $x y^{2}=4 ;(4,1)$