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: Let the coordinates of the point be \(x\) and \(y\).
Given point is
\[ \left(\frac{8t}{1+t^2},\frac{4(1-t^2)}{1+t^2}\right) \] So,
\[ x=\frac{8t}{1+t^2} \] and
\[ y=\frac{4(1-t^2)}{1+t^2} \]

Step 2: Compare with standard parametric form of a circle.
The standard parametric form of a circle is
\[ x=2a\frac{t}{1+t^2} \] and
\[ y=a\frac{1-t^2}{1+t^2} \] This represents the circle
\[ x^2+y^2=a^2 \]

Step 3: Identify the value of radius.
Here,
\[ x=\frac{8t}{1+t^2}=2(4)\frac{t}{1+t^2} \] and
\[ y=4\frac{1-t^2}{1+t^2} \] Therefore,
\[ a=4 \]

Step 4: Verify by direct substitution.
Now,
\[ x^2+y^2 = \left(\frac{8t}{1+t^2}\right)^2 + \left(\frac{4(1-t^2)}{1+t^2}\right)^2 \] \[ = \frac{64t^2+16(1-t^2)^2}{(1+t^2)^2} \] \[ = \frac{16\left[4t^2+(1-t^2)^2\right]}{(1+t^2)^2} \] \[ = \frac{16\left[4t^2+1-2t^2+t^4\right]}{(1+t^2)^2} \] \[ = \frac{16(1+2t^2+t^4)}{(1+t^2)^2} \] \[ = \frac{16(1+t^2)^2}{(1+t^2)^2} \] \[ =16 \] Thus,
\[ x^2+y^2=16 \] \[ x^2+y^2=4^2 \]

Step 5: Final conclusion.
Hence, the point lies on a circle of radius \(4\).
\[ \boxed{\text{Circle of radius }4} \]