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Q.
Let $P(x, y)$ be a variable point on the curve $4 x^2+9 y^2-8 x-36 y+15=0$, then $\min .\left(x^2-2 x+y^2-4 y+5\right)+\max .\left(x^2-2 x+y^2-4 y+5\right)$ is equal to
Conic Sections
Solution:
We have, $\frac{( x -1)^2}{\frac{25}{4}}+\frac{( y -2)^2}{\frac{25}{9}}=1$
So, $\min .\left((x-1)^2+(y-2)^2\right)=\frac{25}{9}$ and $\max \cdot\left((x-1)^2+(y-2)^2\right)=\frac{25}{4}$.