Question

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

• A. $|t|<2]$
• B. $|t| ≤ 1$
• 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 $1-t^{2} \geq 0$ i.e., $|t| \leq 1$