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A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with
 

Option: 1

length of latus rectum 3


Option: 2


length of latus rectum 6


Option: 3

focus\left ( \frac{4}{3},0 \right )


Option: 4

focus\left ( 0,\frac{3}{4} \right )


Answers (1)

Let point \mathrm{p:\left(x_{1}, y_{1}\right)}

equation of tangent at \mathrm{p}:

\mathrm{y-y_{1}=m\left(x-x_{1}\right) }

\mathrm{Q:\left(x_{1}-\frac{y_{1}}{m}, 0\right)}

\because y-axis bisexts the segment PQ

\mathrm{\therefore \frac{x_{1}+x_{1}-\frac{y_{1}}{m}}{2}=0 }

\mathrm{y_{1}=2 m x_{1}}

\mathrm{\therefore\frac{d y}{y}=\frac{d x}{2 x}}

On Integrating both the sides

\mathrm{2 \ln y=\ln x+\ln c}

\mathrm{y^{2}=c x}

passes through \mathrm{(3,3)}

\mathrm{\therefore c=3}

\mathrm{\therefore} equation of parabola: \mathrm{ y^{2}=3 x}

\mathrm{ \therefore }   length of latus rectum = 3

Posted by

Ramraj Saini

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