What is the slope of the normal to the parabo | vedclass

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(B) Given the equation of the parabola is $x^2 + 4y = 0$.We can rewrite this as $4y = -x^2$,which implies $y = -\frac{1}{4}x^2$.To find the slope of t

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$1$

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(B) Given the equation of the parabola is $x^2 + 4y = 0$.We can rewrite this as $4y = -x^2$,which implies $y = -\frac{1}{4}x^2$.To find the slope of the tangent at any point $(x, y)$,we differentiate with respect to $x$:$\frac{dy}{dx} = -\frac{1}{4} 	imes 2x = -\frac{x}{2}$.At the point $(2, -1)$,the slope of the tangent $(m_t)$ is:$m_t = -\frac{2}{2} = -1$.The slope of the normal $(m_n)$ is the negative reciprocal of the slope of the tangent:$m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.Therefore,the slope of the normal is $1$.

What is the slope of the normal to the parabola $x^2 + 4y = 0$ at the point $(2, -1)$?

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