The normal to the curve xy = 4 at the point ( | vedclass

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(B) Given the curve $xy = 4$,we can write $y = \frac{4}{x}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{4}{x^2}$.At the point $

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$(-2,-2)$

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(B) Given the curve $xy = 4$,we can write $y = \frac{4}{x}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{4}{x^2}$.At the point $(2,2)$,the slope of the tangent is $m_t = -\frac{4}{2^2} = -1$.The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{1}{-1} = 1$.The equation of the normal at $(2,2)$ is $y - 2 = 1(x - 2)$,which simplifies to $y = x$.To find the point where the normal meets the curve again,we substitute $y = x$ into the equation of the curve $xy = 4$:$x(x) = 4 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.Since $x = 2$ is the original point,the other point is $x = -2$.Substituting $x = -2$ into $y = x$,we get $y = -2$.Thus,the normal meets the curve again at the point $(-2, -2)$.

The normal to the curve $xy = 4$ at the point $(2,2)$ will meet the curve again at which point?

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