The values of $\lambda$, for which $(\lambda, \lambda-2)$ lies inside the ellipse $4 x^2+9 y^2=36$ and…

The values of $\lambda$, for which $(\lambda, \lambda-2)$ lies inside the ellipse $4 x^2+9 y^2=36$ and outside the parabola $y^2=x$, satisfy
  1. $0 < \lambda < 1$
  2. $0 \leq \lambda \leq 1$
  3. $0 < \lambda < \frac{36}{13}$
  4. $\lambda \in[1,4]$

Solution

The point $(\lambda, \lambda-2)$ lies inside the ellipse $ \begin{aligned} & 4 x^2+9 y^2=36 \\ & \Rightarrow \quad 4 \lambda^2+9(\lambda-2)^2 < 36 \\ & \Rightarrow \quad 4 \lambda^2+9 \lambda^2+36-36 \lambda < 36 \\ & \Rightarrow \quad 13 \lambda^2-36 \lambda < 0 \\ & \Rightarrow \lambda(13 \lambda-36) < 0 \\ & \Rightarrow \quad 0 < \lambda < \frac{36}{13} \end{aligned} $ The point $(\lambda, \lambda-2)$ also lies outside the parabola $y^2=x$ $ \begin{aligned} & (\lambda-2)^2-\lambda>0 \\ \Rightarrow & \lambda^2-4 \lambda+4-\lambda>0 \Rightarrow \lambda^2-5 \lambda+4>0 \\ \Rightarrow & \lambda^2-4 \lambda-\lambda+4>0 \Rightarrow(\lambda-4)(\lambda-1)>0 \\ \Rightarrow & \lambda \in(-\infty, 1) \cup(4, \infty) \end{aligned} $ From Eqs. (i) and (ii), we can conclude that $0 < \lambda < 1$

Asked in: AP EAMCET 2022 (06 Jul Shift 2)

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