Question

The line $y=2t^2$ meets 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. $\left|t\right|<3$
• D. None of these

Answer 1

The correct option is A $\left|t\right|\le 1$

Substitute $y=2t^2$ in the equation of the given ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$,

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

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

This will give real values of $x$ is $1-t^2\ge 0\Rightarrow \left|t\right|\le 1$.