Question

The lines $x=a{t}^2$ meets the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, in the real points, if

• A. $\left|t\right|<2$
• B. $\left|t\right|\le 1$
• C. $\left|t\right|\ge 1$
• D. none of these

Answer 1

The correct option is B

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

Explanation for correct answer:

Given: $x=a{t}^2$ meets $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

If the line meets the ellipse then put $x=a{t}^2$

$$\begin{gathered} \frac{{\left(a{t}^2\right)}^2}{a^2}+\frac{y^2}{b^2}=1 \\ \frac{a^2{t}^4}{a^2}+\frac{y^2}{b^2}=1 \\ {t}^4+\frac{y^2}{b^2}=1 \\ \frac{y^2}{b^2}=1-t^4 \\ \left(1-t^4\right)b^2=y^2 \\ {y}^2=(1-t^2)(1+t^2)b^2 \end{gathered}$$

$(1-t^2)(1+t^2)b^2$ has to be positive as they are equal to square and square is positive, so

$$\begin{gathered} \Rightarrow 1-t^2\ge 0 \\ \Rightarrow 1\ge t^2 \\ \Rightarrow \left|t\right|\le 1 \end{gathered}$$

Hence option B is correct.