Find the equation of the tangent to the curve | vedclass

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(B) Given the curve equation is $y^2 = 6x$.Differentiating both sides with respect to $x$,we get:$2y \frac{dy}{dx} = 6$$\frac{dy}{dx} = \frac{6}{2y} =

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$x + y + 1 = 0$

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(B) Given the curve equation is $y^2 = 6x$.Differentiating both sides with respect to $x$,we get:$2y \frac{dy}{dx} = 6$$\frac{dy}{dx} = \frac{6}{2y} = \frac{3}{y}$Now,find the slope of the tangent at the point $(2, -3)$:$m = \left( \frac{dy}{dx} \right)_{(2, -3)} = \frac{3}{-3} = -1$The equation of the tangent line passing through $(x_1, y_1) = (2, -3)$ with slope $m = -1$ is given by:$y - y_1 = m(x - x_1)$$y - (-3) = -1(x - 2)$$y + 3 = -x + 2$$x + y + 1 = 0$

Find the equation of the tangent to the curve $y^2 = 6x$ at the point $(2, -3)$.

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