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  • Let S be the focus of and a point P is moving on the curve such that it's absci

Let S be the focus of \mathrm{y^2=4 x} and a point P is moving on the curve such that it's abscissa is increasing at the rate of 4 units/sec, then the rate of increase of projection of SP on \mathrm{x+y=1} when P is at \mathrm{(4,4)} is

Option: 1

\sqrt{2}


Option: 2

-1


Option: 3

-\sqrt{2}


Option: 4

-\frac{3}{\sqrt{2}}


Answers (1)

best_answer

       

\mathrm{\overrightarrow{\mathrm{V}}=\left(\mathrm{T}^2-1\right) \hat{\mathrm{i}}+2 \mathrm{~T} \hat{\mathrm{j}}}

\mathrm{\vec{n}=\hat{j}-\hat{i}}

\mathrm{\text { direction of } \vec{V} \text { on } \vec{n}}

\mathrm{\mathrm{y}=\frac{\overrightarrow{\mathrm{V}} \cdot \overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{n}}|}=\frac{\left(1-\mathrm{T}^2\right)+2 \mathrm{~T}}{\sqrt{2}}}

\mathrm{\sqrt{2} \mathrm{y}-1-\mathrm{T}^2+2 \mathrm{~T} ; \sqrt{2} \frac{\mathrm{dy}}{\mathrm{dx}}=-2 \mathrm{~T} \frac{\mathrm{dT}}{\mathrm{dt}}+2 \frac{\mathrm{dT}}{\mathrm{dt}}}

\mathrm{\text { Given } \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{u} \text {; }}                 \mathrm{\text { but } \mathrm{x}=\mathrm{T}^2 ; \, \, \, \quad \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{~T} \frac{\mathrm{dT}}{\mathrm{dt}}}

\mathrm{\text { when } \mathrm{P}(4,4) \text { then } \mathrm{T}=2}        \mathrm{\Rightarrow}       \mathrm{\mathrm{u}=2 \cdot 2 \frac{\mathrm{dT}}{\mathrm{dt}} ; \quad \frac{\mathrm{dT}}{\mathrm{dt}}=1}

\mathrm{\therefore \quad \sqrt{2} \frac{\mathrm{dy}}{\mathrm{dt}}=-4+2=-2}               \mathrm{\Rightarrow}                   \mathrm{\frac{\mathrm{dy}}{\mathrm{dt}}=-\sqrt{2}}

Posted by

Devendra Khairwa

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