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
- $0 < \lambda < 1$
- $0 \leq \lambda \leq 1$
- $0 < \lambda < \frac{36}{13}$
- $\lambda \in[1,4]$
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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