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

The correct option is A $\frac{325}{36}$

$4x^2+9y^2-8x-36y+15=04(x^2-2x+1)+9(y^2-4y+4)-25=04{(x-1)}^2+9{(y-2)}^2=25\frac{{(x-1)}^2}{{\left(\frac{5}{2}\right)}^2}+\frac{{(y-2)}^2}{{\left(\frac{5}{3}\right)}^2}=1$

we have to find min$(x^2-2x+y^2-4y+5)$$+$max$(x^2-2x+y^2-4y+5)$

$=min\{{(x-1)}^2+{(y-2)}^2\}+max\{{(x-1)}^2+{(y-2)}^2\}$

clearly min when$x=1$ and maximum when$y=2$

$min\{{(x-1)}^2+{(y-2)}^2\}+max\{{(x-1)}^2+{(y-2)}^2\}$

$={\left(\frac{5}{3}\right)}^2+{\left(\frac{5}{2}\right)}^2=\frac{325}{36}$

So option $A$ is correct