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

Answer 1

Answer: (B)

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

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., $y^2=b^2(1-t^4)$
$=b^2(1+t^2)(1-t^2)$
$y$ is real, if $1-t^2\ge 0$ i.e., $\left|t\right|\le 1$.