Let P be the point $$(-3$$, $$0)$$ and Q be a moving point (0$$,$$3t). Let PQ be trisected at R so that R is nearer to Q. RN is drawn perpendicular to PQ meeting the $$x-axis$$ at N. The locus of the $$mid-point$$ of RN is

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Infinity Learn · Accepted Answer

Since R is point of trisection R⇒R divides PQ in the ratio $$2$$:$$1$$ $$=2(0)+1(-3)3$$, $$2(3t)+1(0)3$$ $$P($$−$$3$$,$$0)$$,$$Q(0$$,$$3t)$$,$$R($$−$$1$$,$$2t)$$ Let the mid point of RN be (h$$,$$k)R(-1$$,$$2t)  $$+$$ $$N(x$$,$$y)$$ $$2=(h$$,$$k)-1+x=2h$$   $$2t+y=2kN=(2h+1$$,$$0)$$∴    $$k=t$$    RN⊥PQ⇒$$2t$$−$$2h$$−$$2$$×$$3t3=$$−$$1$$⇒    $$2t2=2h+2$$⇒    $$t2=h+1$$⇒    $$k2=h+1$$ Locus     of (h$$,$$k) is $$y2=x+1$$

Let P be the point (-3, 0) and Q be a moving point (0,3t). Let PQ be trisected at R so that R is nearer to Q. RN is drawn perpendicular to PQ meeting the x-axis at N. The locus of the mid-point of RN is

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