UPSEE Solved Model Paper 4

The given Hyperbola is $\frac{x^2}{9} - \frac{y^2}{25}$ = 1 or $\frac{x^2}{9} - \frac{y^2}{25}$ - 1 = 0 Then any point $P(x_1, y_1)$ will lie inside, on or the outside of the hyperbola $\frac{x^2}{9} - \frac{y^2}{25}$ = 1 depending upon the value of $\left( \frac{x_1^2}{9} - \frac{y_1^2}{25} - 1 \right)$ If the value of $\frac{x_1^2}{9} - \frac{y_1^2}{25}$ - 1 > 0 then point $P(x_1, y_1)$ lies inside of the hyperbola If the value of $\frac{x_1^2}{9} - \frac{y_1^2}{25}$ - 1 = 0 then point $P(x_1, y_1)$ lies on the periphery of the hyperbola If the value of $\frac{x_1^2}{9} - \frac{y_1^2}{25}$ - 1 < 0 then point $P(x_1, y_1)$ lies outside of the hyperbola Now let the value of the term $\frac{x_1^9}{\frac{y_1^2}{25}}$ - 1 at given point (2 , - 3) is $\left( \frac{x_1^2}{9} - \frac{y_1^2}{25} - 1 \right)_{(2,-3)}$ Then $\left( \frac{x_1^2}{9} - \frac{y_1^2}{25} - 1 \right)_{(2,-3)}$ = $\frac{2^2}{9} - \frac{(-3)^2}{25} - 1$ ⇒ $\frac{100-81-225}{225}$ = $\frac{-206}{225}$ < 0 As value of term $\frac{x_1^2}{9} - \frac{y_1^2}{25} - 1$ at point P (2 , - 3) is less than 0. Hence the point lies outside of the hyperbola So correct option is B