The line y = 2t^2 intersects the ellipse frac | vedclass

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(A) Given the equation of the ellipse: $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and the line $y = 2t^2$. Substitute $y = 2t^2$ into the ellipse equation: $

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$|t| \leq 1$

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(A) Given the equation of the ellipse: $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and the line $y = 2t^2$. Substitute $y = 2t^2$ into the ellipse equation: $\frac{x^2}{9} + \frac{(2t^2)^2}{4} = 1$ $\frac{x^2}{9} + \frac{4t^4}{4} = 1$ $\frac{x^2}{9} + t^4 = 1$ $x^2 = 9(1 - t^4)$ For the intersection points to be real,$x^2$ must be greater than or equal to $0$: $9(1 - t^4) \geq 0$ $1 - t^4 \geq 0$ $t^4 \leq 1$ $(t^2 - 1)(t^2 + 1) \leq 0$ Since $t^2 + 1 > 0$ for all real $t$,we must have $t^2 - 1 \leq 0$. $t^2 \leq 1$ $|t| \leq 1$

The line $y = 2t^2$ intersects the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ in real points if

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