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A line which makes an acute angle \mathrm{\Theta } with the positive direction of the \mathrm{ x-axis } is drawn through the point \mathrm{ P(3, 4) }, to cut the curve \mathrm{ y2 = 4x} at \mathrm{ Q}  and \mathrm{R}. The lengths of the segments \mathrm{PQ}  and \mathrm{PR} are numerical values of the roots of the equation \mathrm{r2} \mathrm{sin2} \mathrm{\Theta }  \mathrm{ + 4r } \mathrm{\left ( 2 sin \Theta - cos \Theta \right )} \mathrm{+ c } \mathrm{= 0, } where \mathrm{c = }

 

Option: 1

\mathrm{4}


Option: 2

\mathrm{3}


Option: 3

\mathrm{5}


Option: 4

\mathrm{2}


Answers (1)

best_answer

Equation of the line passing through \mathrm{P(3, 4) } and making an angle \mathrm{\Theta } with the positive direction of the \mathrm{x-axis } is \mathrm{\frac{x-3}{cos \Theta }=\frac{y-4}{sin \Theta } =r\ \ \ ...........\left ( 1 \right )} 
 

where r is the distance of any point on the line from the point \mathrm {P}. The coordinates of any point on \mathrm {(i)}  are 

\mathrm {(r\ cos \Theta \ + 3,r \ sin \Theta \ +4)} . Line \mathrm { (1) } cuts the curve \mathrm {y2 = 4x}  at the points \mathrm {Q} and \mathrm {R} , whose distances \mathrm {PQ}  and \mathrm {PR } from the point \mathrm {P} are given by \mathrm {(4 \ +\ r \ sin\Theta )2=4(3+\ r\ cos\Theta )}

or \mathrm {r2}  \mathrm {sin2\ \Theta \ +\ 4r(2\ sin\ \Theta -\ cos\ \Theta )\ +\ 4\ =\ 0}

 

Posted by

Ritika Kankaria

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