Q. The line $x = at^{2}$ meets the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,$ in the real points, iff
A) $|t| < 2$
B) $|t| \leq 1$
C) $|t| > 1$
D) None of these
Explanation
(b)
Put $x = at^{2}$ in the given equation, we get
$\frac{a^{2}t^{4}}{a^{2}} + \frac{y^{2}}{{\ b}^{2}} = 1$
$y^{2} = b^{2}\left( 1 - t^{2} \right)(1 + t^{2})$
This will give real values of $y,$ if
$1 - t^{2} \geq 0\ \Rightarrow \ |t| \leq 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.