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Grade 12th passAnalytical Geometry

  • The locus of the point represented by x= t^2 +t+1 and y=t^-t+1 is

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4 Years agoGrade 12th pass
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ApprovedApproved Tutor Answer1 Year ago

To find the locus of the point represented by the equations \( x = t^2 + t + 1 \) and \( y = t^{-t} + 1 \), we need to eliminate the parameter \( t \) and express \( y \) in terms of \( x \). This process will help us understand the relationship between \( x \) and \( y \) as \( t \) varies.

Step 1: Analyze the Equation for x

The equation for \( x \) is a quadratic function of \( t \):

x = t^2 + t + 1

This can be rewritten in standard form as:

x = (t + 0.5)^2 + 0.75

This indicates that \( x \) is always greater than or equal to \( 0.75 \), since the square of any real number is non-negative.

Step 2: Investigate the Equation for y

The equation for \( y \) is more complex:

y = t^{-t} + 1

Here, \( t^{-t} \) represents an exponential function where the base is a variable. The behavior of this function can be tricky, as it depends on the value of \( t \). For positive values of \( t \), \( t^{-t} \) decreases rapidly as \( t \) increases, while for negative values, it can behave quite differently.

Step 3: Finding the Relationship

To find a relationship between \( x \) and \( y \), we can express \( t \) in terms of \( x \) from the first equation:

  • Rearranging gives us a quadratic equation: \( t^2 + t + (1 - x) = 0 \).
  • Using the quadratic formula, we can solve for \( t \):
  • t = \frac{-1 \pm \sqrt{1 - 4(1 - x)}}{2}

Next, we substitute this expression for \( t \) back into the equation for \( y \). However, since \( y = t^{-t} + 1 \) is quite complex, we may not arrive at a simple expression for \( y \) in terms of \( x \) directly.

Step 4: Behavior of y as t Changes

To understand the behavior of \( y \), we can analyze specific values of \( t \):

  • If \( t = 0 \), then \( y = 1 + 1 = 2\).
  • If \( t = 1 \), then \( y = 1 + 1 = 2\).
  • If \( t = 2 \), then \( y \) decreases significantly.

This suggests that \( y \) has a minimum value at certain points and may not have a straightforward relationship with \( x \). The function \( t^{-t} \) approaches 0 as \( t \) increases, indicating that \( y \) will approach 1 for large values of \( t \).

Final Thoughts

In summary, while we can express \( x \) in terms of \( t \) and understand the behavior of \( y \), finding a direct relationship between \( x \) and \( y \) is complicated due to the nature of \( t^{-t} \). The locus of the point is not a simple geometric shape but rather a more complex curve that depends on the interplay of these two functions. Further analysis or numerical methods might be required to visualize this relationship fully.