Question

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

• A. $|t|<2]$
• B. $|t|\le 1$
• C. $|t|>1$
• D. None of these

Answer 1

Answer: (B)

Putting $x=a{t}^2$ in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we get
$t^4+\frac{y^2}{b^2}=1$
$i.e.,y^2=b^2\left(1-t^4\right)$
$=b^2\left(1+t^2\right)\left(1-t^2\right)$
$y$ is real, if $1-t^2\ge 0$ i.e., $|t|\le 1$