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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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