10-2. Parabola, Ellipse, Hyperbola
hard

અહી $S=\left\{(x, y) \in N \times N : 9(x-3)^{2}+16(y-4)^{2} \leq 144\right\}$ અને $\quad T=\left\{(x, y) \in R \times R :(x-7)^{2}+(y-4)^{2} \leq 36\right\}$ હોય તો $n ( S \cap T )$ ની કિમંત $......$ થાય.

A

$27$

B

$26$

C

$25$

D

$24$

(JEE MAIN-2022)

Solution

$S: \frac{(x-3)^{2}}{16}+\frac{(y-4)^{2}}{9} \leq 1 ; x, y \in\{1,2,3, \ldots \ldots\}$

$T:(x-7)^{2}+(y-4)^{2} \leq 36 x, y \in R$

Let $x-3=x: y-4=y$

$S: \frac{x^{2}}{16}+\frac{y^{2}}{9} \leq 1 ; x \in\{-2,-1,0,1, \ldots \ldots\}$

$T:(x-4)^{2}+y^{2} \leq 36 ; y \in\{-3,-2,-1,0, \ldots \ldots\}$

$S \cap T =(-2,0),(-1,0), \ldots .(4,0) \rightarrow(7)$

$(-1,1),(0,1), \ldots \ldots(3,1) \rightarrow(5)$

$(-1,-1),(0,-1), \ldots \ldots(3,-1) \rightarrow(5)$

$(-1,2),(0,2),(1,2),(2,2) \rightarrow(4)$

$(-1,-2),(0,-2),(1,-2),(2,-2) \rightarrow(4)$

$(0,3)(0,-3) \rightarrow(2)$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.