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PQ is a focal chord of the parabola \mathrm{y^2=4 a x}, O is the origin. Find the coordinates of the centroid, G, of triangle OPQ and hence find the locus of G as PQ varies.

 

Option: 1

circle


Option: 2

straight line 


Option: 3

ellipse


Option: 4

parabola


Answers (1)

best_answer

Let P be \mathrm{\left(a t_1^2, 2 a t_1\right)} . Then Q is \mathrm{\left(a t_2^2, 2 a t_2\right)}  with the condition\mathrm{t_1 t_2=-1} (condition for focal chord).

∴ if G (x, y) be the centroid of ΔOPQ, then \mathrm{\left.\begin{array}{l} x=\frac{a t_1^2+a t_2^2}{3} \\ y=\frac{2 a\left(t_1+t_2\right)}{3} \end{array}\right\} \text { with } t_1 t_2=-1}

\mathrm{\text { Eliminating } t_1 \text { and } t_2 \text { between } x \text { and } y \text {, we have }}

\mathrm{\begin{aligned} & y^2=\frac{4 a^2}{9}\left[t_1^2+t_2^2+2 t_1 t_2\right]=\frac{4 a}{3} \cdot \frac{a}{3}\left(t_1^2+t_2^2\right)+\frac{8 a^2}{9} t_1 t_2 \\= & \left.\frac{4 a}{3} \cdot x-\frac{8 a^2}{9} \text { (using } t_1 t_2=-1\right) \end{aligned}}

\mathrm{\therefore \text { the locus of } G \text { (as } t_1 \text { varies) is } 9 y^2=12 a x-8 a^2}       \mathrm{\text { i.e., } 9 y^2=12 a\left(x-\frac{2}{3} a\right)}

which is a parabola with vertex at \mathrm{\left(\frac{2 a}{3}, 0\right)}  and latus rectum of length \mathrm{\frac{12 a}{9}}

 

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Gaurav

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