Question:
For the parabola represented in the parametric form
\[
x=t^2+t+1
\]
\[
y=t^2-t+1,
\]
the length of latus rectum is:
For the parabola represented in the parametric form
\[
x=t^2+t+1
\]
\[
y=t^2-t+1,
\]
the length of latus rectum is:
Show Hint
Convert the parametric equations into a Cartesian equation and compare with the standard parabola
\[
X^2=4aY.
\]
The latus rectum length is always \(4a\).
Updated On: Jun 26, 2026
- \(2\)
- \(3\)
- \(\frac12\)
- \(8\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Eliminate the parameter.
Given \[ x=t^2+t+1, \] \[ y=t^2-t+1. \] Subtracting, \[ x-y=2t. \] Hence, \[ t=\frac{x-y}{2}. \] Adding, \[ x+y=2t^2+2. \] Thus, \[ x+y-2=2t^2. \] Substituting \(t=\frac{x-y}{2}\), \[ x+y-2=\frac{(x-y)^2}{2}. \]
Step 2: Transform to standard form.
Let \[ X=x-y,\qquad Y=x+y-2. \] Then \[ X^2=2Y. \] Comparing with \[ X^2=4aY, \] we get \[ 4a=2. \] Hence, \[ a=\frac12. \]
Step 3: Find the length of latus rectum.
Length of latus rectum \[ =4a =4\left(\frac12\right) =2. \]
Step 4: Final conclusion.
Therefore, \[ \boxed{2}. \]
Given \[ x=t^2+t+1, \] \[ y=t^2-t+1. \] Subtracting, \[ x-y=2t. \] Hence, \[ t=\frac{x-y}{2}. \] Adding, \[ x+y=2t^2+2. \] Thus, \[ x+y-2=2t^2. \] Substituting \(t=\frac{x-y}{2}\), \[ x+y-2=\frac{(x-y)^2}{2}. \]
Step 2: Transform to standard form.
Let \[ X=x-y,\qquad Y=x+y-2. \] Then \[ X^2=2Y. \] Comparing with \[ X^2=4aY, \] we get \[ 4a=2. \] Hence, \[ a=\frac12. \]
Step 3: Find the length of latus rectum.
Length of latus rectum \[ =4a =4\left(\frac12\right) =2. \]
Step 4: Final conclusion.
Therefore, \[ \boxed{2}. \]
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