Question

For any real number $t$, the point $\left( \dfrac{8t}{1 + t^2}, \dfrac{4(1 - t^2)}{1 + t^2} \right)$ 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

Hint:

Convert parametric form to Cartesian using identity, then test equation type (circle, ellipse).

Answer 1

The Correct Option is B

Let $x = \dfrac{8t}{1 + t^2}$ and $y = \dfrac{4(1 - t^2)}{1 + t^2}$
We try: $x^2 + y^2 = \left( \dfrac{8t}{1 + t^2} \right)^2 + \left( \dfrac{4(1 - t^2)}{1 + t^2} \right)^2$
$= \dfrac{64t^2 + 16(1 - 2t^2 + t^4)}{(1 + t^2)^2}$
$= \dfrac{64t^2 + 16(1 - 2t^2 + t^4)}{(1 + t^2)^2} = \dfrac{64t^2 + 16 - 32t^2 + 16t^4}{(1 + t^2)^2}$
$= \dfrac{32t^2 + 16 + 16t^4}{(1 + t^2)^2} = \dfrac{16(t^4 + 2t^2 + 1)}{(1 + t^2)^2} = \dfrac{16(1 + t^2)^2}{(1 + t^2)^2} = 16$
Thus, $x^2 + y^2 = 16 \Rightarrow$ circle of radius $4$