For any real number t, the point ( 8t1+t^2 , 4(1-t^2)1+t^2 ) lies on a/ an on a/an

Question

Accepted Answer

Correct answer: B. (a) x= 8t1+t^2 ,y= 4(1-t^2)(1+t^2)x^2+y^2= 64t^2(1+t^2)^2+ 16(1+t^2)^2(1+t^2)^2 = 64t^2+16(1-t^2)^2(1+t^2)^2= 64t^2+16+16t^4-32t^2(1+t^2)^2= 16t^4+16+32t^2(1+t^2)^2= 16(1+t^2)^2(1+t^2)^2=16 ∵x^2+y^2=16= (4 )^2 ∴ (x ,y ) lies on circle with radius 4

$For any real number t, the point (\frac{8 t}{1 + t ^{2}}, \frac{4 ( 1 - t ^{2} )}{1 + t ^{2}}) lies on a / an on a/an$

Solution

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