Question

The values of \(\lambda\), for which \((\lambda,\lambda-2)\) lies inside the ellipse \[ 4x^2+9y^2=36 \] and outside the parabola \[ y^2=x, \] satisfy

• A. \(0\lt \lambda\lt 1\)
• B. \(0\leq \lambda \leq 1\)
• C. \(0\lt \lambda\lt \frac{36}{13}\)
• D. \(\lambda \notin [1,4]\)

Hint:

For a point to lie inside an ellipse, substitute the point and use the inequality form. For a point to lie outside a parabola \(y^2=x\), compare \(y^2\) with \(x\) carefully.

Answer 1

The Correct Option is A

Step 1: Use the condition for the point to lie inside the ellipse.
The given point is \[ (\lambda,\lambda-2) \] The ellipse is \[ 4x^2+9y^2=36 \] For the point to lie inside the ellipse, \[ 4x^2+9y^2\lt 36 \] Substituting \[ x=\lambda,\quad y=\lambda-2 \] we get \[ 4\lambda^2+9(\lambda-2)^2\lt 36 \] Expanding, \[ 4\lambda^2+9(\lambda^2-4\lambda+4)\lt 36 \] \[ 4\lambda^2+9\lambda^2-36\lambda+36\lt 36 \] \[ 13\lambda^2-36\lambda\lt 0 \] \[ \lambda(13\lambda-36)\lt 0 \] Therefore, \[ 0\lt \lambda\lt \frac{36}{13} \]

Step 2: Use the condition for the point to lie outside the parabola.
The parabola is \[ y^2=x \] For the point \[ (\lambda,\lambda-2) \] we compare \[ (\lambda-2)^2 \] with \[ \lambda \] To be outside the parabola, \[ (\lambda-2)^2\gt \lambda \] Now, \[ \lambda^2-4\lambda+4\gt \lambda \] \[ \lambda^2-5\lambda+4\gt 0 \] Factorizing, \[ (\lambda-1)(\lambda-4)\gt 0 \] Thus, \[ \lambda\lt 1 \quad \text{or} \quad \lambda\gt 4 \]

Step 3: Find the common values of \(\lambda\).
From the ellipse condition, \[ 0\lt \lambda\lt \frac{36}{13} \] From the parabola condition, \[ \lambda\lt 1 \quad \text{or} \quad \lambda\gt 4 \] Taking common values, \[ 0\lt \lambda\lt 1 \]

Step 4: Final conclusion.
Hence, the required values of \(\lambda\) are \[ \boxed{0\lt \lambda\lt 1} \]