For any real number t, the point 8t1+t2,41-t21+t2 lies on a/an

(C) Ellipse with 4 as its major axis length

(D) Ellipse with 4 as its minor axis length

AP EAMCET · Maths · Circle

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

A Circle of radius $2$
B Circle of radius $4$
C Ellipse with $4$ as its major axis length
D Ellipse with $4$ as its minor axis length

Verified Solution

Answer & Solution

Correct Answer

(B) Circle of radius $4$

Step-by-step Solution

Detailed explanation

Let, x=8t1+t2 .......1 And y=41-t21+t2 ......2 Now squaring and adding both the equation we get, x2+y2=64t2+161-t221+t22 ⇒x2+y2=161-t22+4t21+t22 ⇒x2+y2=161+t221+t22 ⇒x2+y2=16 Hence point lie on the circle of radius 4

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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