Question

Let $(x,y)$ be a variable point on the curve $4x^2+9y^2-8x-36y+15=0$ Then, min $(x^2-2x+y^2-4y+5)$ +max $(x^2-2x+y^2-4y+5)$ is

• A. $\frac{325}{36}$
• B. $\frac{36}{325}$
• C. $\frac{13}{25}$
• D. $\frac{25}{13}$

Answer 1

Answer: (A)

We have,
$4x^2+9y^2-8x-36y+15=0$
$\Rightarrow 4(x^2-2x+1)+9(y^2-4y+4)$
$=-15+4+36$
$\Rightarrow 4(x-1{)}^2+9(y-2{)}^2=25$
$\Rightarrow \frac{{\left(x-1\right)}^2}{25/4}+\frac{{\left(y-2\right)}^2}{25/9}=1$
Now, $\min (x^2-2x+y^2-4y+5)+\max (x^2-2x+y^2-4y+5)$
$=\min [(x-1{)}^2+(y-2{)}^2]+\max [(x-1{)}^2+(y-2{)}^2]$
$=\frac{25}{9}+\frac{25}{4}$
$=25\left(\frac{4+9}{36}\right)$
$=\frac{25\times 13}{36}$
$=\frac{325}{36}$