Question

The line y = 2t² meets the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ in real points if

• A. |t| ≤ 1
• B. |t| > 1
• C. |t| < 3
• D. None of these

Answer 1

Answer: (A)

|t| ≤ 1

Answer 2

Putting y = 2t² in the equation of the given ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$,

we get $\frac{x^{2}}{9} + \frac{4t^{4}}{4}$ = 1 ⇒ x² = 9 (1 - t⁴) = 9 (1 - t²) (1 + t²).

This will give real value of x if 1 - t² ≥ 0 i.e. |t| ≤ 1.