The line `x = at^(2)` meets the ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1` in the real points iff

Options

A. `|t| lt 2`
B. `|t| le 1`
C. `|t| gt t`
D. none of these

Explanation

To solve the problem, we need to determine the conditions under which the line \( x = at^2 \) intersects the ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) at real points.

### Step-by-Step Solution:

1. **Substituting the Line Equation into the Ellipse Equation**:
We start by substituting \( x = at^2 \) into the ellipse equation:
\[ \frac{(at^2)^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Simplifying this gives:
\[ \frac{a^2 t^4}{a^2} + \frac{y^2}{b^2} = 1 \]
This simplifies to:
\[ t^4 + \frac{y^2}{b^2} = 1 \]

2. **Isolating \( y^2 \)**:
Rearranging the equation to isolate \( y^2 \):
\[ \frac{y^2}{b^2} = 1 - t^4 \]
Multiplying both sides by \( b^2 \):
\[ y^2 = b^2(1 - t^4) \]

3. **Finding Conditions for Real Points**:
For \( y^2 \) to be non-negative (since \( y^2 \geq 0 \)), we need:
\[ b^2(1 - t^4) \geq 0 \]
Since \( b^2 > 0 \) (as \( b \) is a real number), we can simplify this to:
\[ 1 - t^4 \geq 0 \]
This leads to:
\[ t^4 \leq 1 \]

4. **Solving the Inequality**:
Taking the fourth root of both sides, we find:
\[ |t| \leq 1 \]
This can also be expressed as:
\[ -1 \leq t \leq 1 \]

5. **Conclusion**:
Therefore, the line \( x = at^2 \) meets the ellipse in real points if:
\[ |t| \leq 1 \]

### Final Answer:
The line \( x = at^2 \) meets the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) in real points if \( |t| \leq 1 \).

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