If the equation of the normal to the curve x= | vedclass

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(D) Given the parametric equations $x=\sqrt{t}$ and $y=t-\frac{1}{\sqrt{t}}$.At $t=4$,$x=\sqrt{4}=2$ and $y=4-\frac{1}{\sqrt{4}}=4-\frac{1}{2}=\frac{7

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$8x+34y=135$

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(D) Given the parametric equations $x=\sqrt{t}$ and $y=t-\frac{1}{\sqrt{t}}$.At $t=4$,$x=\sqrt{4}=2$ and $y=4-\frac{1}{\sqrt{4}}=4-\frac{1}{2}=\frac{7}{2}$.We find the derivative $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.$\frac{dx}{dt} = \frac{1}{2\sqrt{t}}$ and $\frac{dy}{dt} = 1 + \frac{1}{2t\sqrt{t}}$.At $t=4$,$\frac{dx}{dt} = \frac{1}{4}$ and $\frac{dy}{dt} = 1 + \frac{1}{16} = \frac{17}{16}$.Thus,the slope of the tangent $m_T = \frac{17/16}{1/4} = \frac{17}{4}$.The slope of the normal $m_N = -\frac{1}{m_T} = -\frac{4}{17}$.The equation of the normal is $y - \frac{7}{2} = -\frac{4}{17}(x - 2)$.Multiplying by $34$,we get $34y - 119 = -8(x - 2) \implies 34y - 119 = -8x + 16$.Rearranging gives $8x + 34y = 135$.

If the equation of the normal to the curve $x=\sqrt{t}$,$y=t-\frac{1}{\sqrt{t}}$ at $t=4$ is

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