Q44 / 160 · WBJEE 2009 · Mathematics

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

  • Correct answer
    A
    $|t| \leq 1$
  • B
    $|t| < 1$
  • C
    $|t|>1$
  • D
    $|t| \geq 1$

Solution

$$ \begin{aligned} & \text { Hints: } \frac{x^2}{9}+\frac{y^2}{4}=1 ; \mathrm{y}=2 \mathrm{t}^2 \\ & \frac{x^2}{9}+\frac{4 t^4}{4}=1 \Rightarrow \frac{x^2}{9}+t^4=1 \Rightarrow x^2=9\left(1-t^4\right) \\ & x^2 \geq 0 \Rightarrow 9\left(1-t^4\right) \geq 0 \Rightarrow t^4-1 \leq 0 \\ & \Rightarrow\left(t^2-1\right)\left(t^2+1\right) \leq 0 \\ & \Rightarrow t^2-1 \leq 0 \quad\left(\because t^2+1>0\right) \\ & \Rightarrow|t| \leq 1 \end{aligned} $$

Hence, the correct answer is option: 'A'.

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