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Let P\left ( 4,-4 \right )\: \: and\: \: Q\left ( 9,6 \right ) be two points on the parabola, y^{2}=4x and let X be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of \Delta PXQ is maximum. Then this maximum area (in sq. units) is : 

  • Option 1)

    \frac{625}{4}

     

     

     

  • Option 2)

    \frac{75}{2}

  • Option 3)

    \frac{125}{2}

  • Option 4)

    \frac{125}{4}

Answers (1)

best_answer

 

Standard equation of parabola -

y^{2}=4ax

- wherein

 

 

Parametric coordinates of parabola -

x= at^{2}

y= 2at

- wherein

For the parabola.

y^{2}=4ax

 

m_{ap} = \frac{6 + 4}{9-4} = 2 \\\\ \Rightarrow2yy'= 4 \;\;\;(\textup{for\;maximum\;area}) \\\\\Rightarrow 2y\times 2 = 4 \\\\\Rightarrow y = 1, x = \frac{1}{4} \\\\ R \equiv \left (\frac{1}{4}, 1 \right ) \\\\ \therefore \textup{area} = \frac{125}{4} \;\textup{sq.units}


Option 1)

\frac{625}{4}

 

 

 

Option 2)

\frac{75}{2}

Option 3)

\frac{125}{2}

Option 4)

\frac{125}{4}

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