The line `y=2t^2` meets the ellipse `(x^2)/(9)+(y^2)/(4)=1` in real points if

Options

A. `|t| le 1`
B. `|t| ge 1`
C. `|t| lt 3`
D. None of these

Answer

To determine the conditions under which the line \( y = 2t^2 \) meets the ellipse \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \) in real points, we can follow these steps:

### Step 1: Substitute the line equation into the ellipse equation
We start with the ellipse equation:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
Substituting \( y = 2t^2 \) into the ellipse equation gives:
\[ \frac{x^2}{9} + \frac{(2t^2)^2}{4} = 1 \]

### Step 2: Simplify the equation
Now, simplify the equation:
\[ \frac{x^2}{9} + \frac{4t^4}{4} = 1 \]
This simplifies to:
\[ \frac{x^2}{9} + t^4 = 1 \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ \frac{x^2}{9} = 1 - t^4 \]
Multiplying through by 9 results in:
\[ x^2 = 9(1 - t^4) \]

### Step 4: Analyze the conditions for real points
For \( x^2 \) to be non-negative (since \( x^2 \geq 0 \)), we need:
\[ 9(1 - t^4) \geq 0 \]
This simplifies to:
\[ 1 - t^4 \geq 0 \]
Thus, we have:
\[ t^4 \leq 1 \]

### Step 5: Solve for \( t \)
Taking the fourth root gives:
\[ |t| \leq 1 \]
This means:
\[ -1 \leq t \leq 1 \]

### Conclusion
The line \( y = 2t^2 \) meets the ellipse in real points if:
\[ |t| < 1 \]