If four point t i - 1 - t i i-1-2-3-4 are concyclic then t1t2t3t4-

The correct option is A 1 equation of the circle #160; x^2+y^2+2gx+2fy+c=0 Let x= to amp; y=1/ti ; (ti)^2+ (1ti)^2+2g(ti)+2f (1ti)+ ...

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Standard XII

Mathematics

Chord of Contact: Ellipse

If four point...

Question

If four point $(t_{i}, \frac{1}{t_{i}}) (i = 1, 2, 3, 4)$ are concyclic then $t_{1} t_{2} t_{3} t_{4} =$

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Solution

The correct option is A $1$
equation of the circle
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
Let $x =$ to & $y = 1 / t i$ ;
$(t i)^{2} + {(\frac{1}{t i})}^{2} + 2 g (t i) + 2 f (\frac{1}{t i}) + c = 0$
$(t i)^{4} + 1 + 2 g (t i)^{3} + 2 f (t i) + c = 0$
$(t i)^{4} + 2 g (t i)^{3} + 2 f (t i) + t i c + t i = 0$
$t_{1} t_{2} t_{3} t_{4} = 1$
$\frac{1}{t_{1} t_{2} t_{3} t_{4}} = 1 \Rightarrow 1 = t_{1} t_{2} t_{3} t_{4}$
Option $(A)$ is correct

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If four point (ti,1ti)(i=1,2,3,4) are concyclic then t1t2t3t4=

The correct option is A 1equation of the circle x2+y2+2gx+2fy+c=0Let x= to & y=1/ti;(ti)2+(1ti)2+2g(ti)+2f(1ti)+c=0(ti)4+1+2g(ti)3+2f(ti)+c=0(ti)4+2g(ti)3+2f(ti)+tic+ti=0t1t2t3t4=11t1t2t3t4=1⇒1=t1t2t3t4Option (A) is correct

If four point $(t_{i}, \frac{1}{t_{i}}) (i = 1, 2, 3, 4)$ are concyclic then $t_{1} t_{2} t_{3} t_{4} =$