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, $\min (x^{2}-2x+y^{2}-4y+5)+\max (x^{2}-2x+y^{2}-4y+5)$
$=\min [(x-1)^{2}+(y-2)^{2}]+\max [(x-1)^{2}+(y-2)^{2}]$
$=\frac{25}{9}+\frac{25}{4}$
$=25 \left(\frac{4+9}{36}\right)$
$=\frac{25 \times 13}{36}$
$=\frac{325}{36}$