Let $$P(x$$, $$y)$$ be a variable point such that $$(x$$−$$3)2+(y$$−$$2)2$$−$$(x$$−$$6)2+(y+2)2=3$$. $$Statement-1$$: P traces a hyperbola whose eccentricity $$is53$$.$$Statement-2$$: P traces a hyperbola such that the equation of its conjugate axis is $$6x$$ – $$8y$$ $$=$$ $$27$$

Question

Let $$P(x$$, $$y)$$ be a variable point such that $$(x$$−$$3)2+(y$$−$$2)2$$−$$(x$$−$$6)2+(y+2)2=3$$. $$Statement-1$$: P traces a hyperbola whose eccentricity $$is53$$.$$Statement-2$$: P traces a hyperbola such that the equation of its conjugate axis is $$6x$$ – $$8y$$ $$=$$ $$27$$

Infinity Learn · Accepted Answer

Distance of P from the fixed points $$F1(3$$, $$2)$$ and $$F2(6$$,−$$2)$$ is constant $$3$$.So by definition, locus of P is a hyperbola with foci  $$F1$$ and $$F2$$, length of the transverse axes is $$3$$. If e is the eccentricity then $$F1F2=2ae$$ where $$2a=3$$. ⇒$$e=139+16=53$$ and the $$statement-1$$ is true.n $$statement-2$$ slope of the conjugate axis is −$$6$$−$$3$$−$$2$$−$$2=34$$ and it passes through the $$mid-point$$ of $$F1F2$$ So its equation $$isy=34x$$−$$92$$⇒$$6x$$−$$8y=27Thus$$ $$statement-2$$ is also true but does not lead to $$statement-1$$.

Let P(x, y) be a variable point such that (x−3)2+(y−2)2−(x−6)2+(y+2)2=3. Statement-1: P traces a hyperbola whose eccentricity is53.Statement-2: P traces a hyperbola such that the equation of its conjugate axis is 6x – 8y = 27

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