Question

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

• A. $\left|t\right|\le 1$
• B. $\left|t\right|<1$
• C. $\left|t\right|>1$
• D. $\left|t\right|\ge 1$

Answer 1

Answer: (A)

$\left|t\right|\le 1$

The intersection of line $y=2t^2$ and the ellipse

$\frac{x^2}{9}+\frac{y^2}{4}=1$

$\frac{x^2}{9}+\frac{4t^4}{4}=1$

$\Rightarrow x^2=9(1-t^4)$

Since, $x$ is real, therefore

$9(1-t^4)\ge 0$

$\Rightarrow (1-t^2)(1+t^2)\ge 0$

$\Rightarrow (1-t^2)\ge 0$

$\Rightarrow t^2\le 1$

$\Rightarrow \left|t\right|\le 1$