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:

The parametric equations \(x=\dfrac{2at}{1+t^2}\) and \(y=\dfrac{a(1-t^2)}{1+t^2}\) represent the circle \(x^2+y^2=a^2\).

Answer 1

The Correct Option is B

Step 1: Name the coordinates.
Let $x = \dfrac{8t}{1+t^2}$ and $y = \dfrac{4(1-t^2)}{1+t^2}$. We want to eliminate $t$ and find the curve these points trace.
Step 2: Recall the classic half-angle substitution.
If $t = \tan\frac{\theta}{2}$, then $\dfrac{2t}{1+t^2} = \sin\theta$ and $\dfrac{1-t^2}{1+t^2} = \cos\theta$. These are exactly the pieces appearing here.
Step 3: Rewrite $x$ and $y$ with $\theta$.
We have $x = \dfrac{8t}{1+t^2} = 4\cdot\dfrac{2t}{1+t^2} = 4\sin\theta$, and $y = 4\cdot\dfrac{1-t^2}{1+t^2} = 4\cos\theta$.
Step 4: Form $x^2+y^2$.
Then \[ x^2 + y^2 = 16\sin^2\theta + 16\cos^2\theta = 16(\sin^2\theta+\cos^2\theta) = 16. \]
Step 5: Identify the curve.
The relation $x^2+y^2 = 16$ is a circle centred at the origin with radius $\sqrt{16}=4$.
Step 6: State the conclusion.
Hence every such point lies on a circle of radius $4$.
\[ \boxed{\text{Circle of radius } 4} \]