Let S=left{(x, y) in N times N : 9(x-3)^{2}+1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The set $S$ represents points $(x, y)$ with $x, y \in N$ (natural numbers) inside or on the ellipse $\frac{(x-3)^2}{16} + \frac{(y-4)^2}{9} \leq 1

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(B) The set $S$ represents points $(x, y)$ with $x, y \in N$ (natural numbers) inside or on the ellipse $\frac{(x-3)^2}{16} + \frac{(y-4)^2}{9} \leq 1$.Since $x, y \geq 1$,we test integer values for $x$ and $y$:For $x=1: 9(-2)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 36 + 16(y-4)^2 \leq 144$ $\Rightarrow 16(y-4)^2 \leq 108$ $\Rightarrow (y-4)^2 \leq 6.75$. Possible $y \in \{1, 2, 3, 4, 5, 6\}$.For $x=2: 9(-1)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 9 + 16(y-4)^2 \leq 144$ $\Rightarrow 16(y-4)^2 \leq 135$ $\Rightarrow (y-4)^2 \leq 8.43$. Possible $y \in \{1, 2, 3, 4, 5, 6\}$.For $x=3: 9(0)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 16(y-4)^2 \leq 144$ $\Rightarrow (y-4)^2 \leq 9$. Possible $y \in \{1, 2, 3, 4, 5, 6, 7\}$.For $x=4: 9(1)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 9 + 16(y-4)^2 \leq 144$ $\Rightarrow (y-4)^2 \leq 8.43$. Possible $y \in \{1, 2, 3, 4, 5, 6\}$.For $x=5: 9(2)^2 + 16(y-4)^2 \leq 144$ $\Rightarrow 36 + 16(y-4)^2 \leq 144$ $\Rightarrow (y-4)^2 \leq 6.75$. Possible $y \in \{1, 2, 3, 4, 5, 6\}$.Now check which of these points satisfy $T: (x-7)^2 + (y-4)^2 \leq 36$.Points $(x, y)$ in $S$ are: $(1, 1..6), (2, 1..6), (3, 1..7), (4, 1..6), (5, 1..6)$.Checking $(x-7)^2 + (y-4)^2 \leq 36$ for these points:For $x=1: (-6)^2 + (y-4)^2 \leq 36$ $\Rightarrow 36 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 0$ $\Rightarrow y=4$. Point $(1, 4)$.For $x=2: (-5)^2 + (y-4)^2 \leq 36$ $\Rightarrow 25 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 11$ $\Rightarrow y \in \{1, 2, 3, 4, 5, 6, 7\}$. Intersection with $S$ gives $y \in \{1, 2, 3, 4, 5, 6\}$. ($6$ points)For $x=3: (-4)^2 + (y-4)^2 \leq 36$ $\Rightarrow 16 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 20$ $\Rightarrow y \in \{0..8\}$. Intersection with $S$ gives $y \in \{1, 2, 3, 4, 5, 6, 7\}$. ($7$ points)For $x=4: (-3)^2 + (y-4)^2 \leq 36$ $\Rightarrow 9 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 27$ $\Rightarrow y \in \{-1..9\}$. Intersection with $S$ gives $y \in \{1, 2, 3, 4, 5, 6\}$. ($6$ points)For $x=5: (-2)^2 + (y-4)^2 \leq 36$ $\Rightarrow 4 + (y-4)^2 \leq 36$ $\Rightarrow (y-4)^2 \leq 32$ $\Rightarrow y \in \{-1..9\}$. Intersection with $S$ gives $y \in \{1, 2, 3, 4, 5, 6\}$. ($6$ points)Total points $= 1 + 6 + 7 + 6 + 6 = 26$.

Let $S=\left\{(x, y) \in N \times N : 9(x-3)^{2}+16(y-4)^{2} \leq 144\right\}$ and $T=\left\{(x, y) \in R \times R :(x-7)^{2}+(y-4)^{2} \leq 36\right\}$. Then $n(S \cap T)$ is equal to $......$

Similar Questions

The eccentricity of the ellipse $4x^2 + 9y^2 + 8x + 36y + 4 = 0$ is

The extremities of the latera recta of the ellipses $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ having a given major axis $2a$ lie on:

Let a tangent be drawn to the ellipse $\frac{x^{2}}{27}+y^{2}=1$ at $(3 \sqrt{3} \cos \theta, \sin \theta)$ where $\theta \in\left(0, \frac{\pi}{2}\right)$. Then the value of $\theta$ such that the sum of intercepts on axes made by this tangent is minimum is equal to ..... .

The equations $x = a \cos \theta$ and $y = b \sin \theta$ with $a > b$ represent a conic section. Its eccentricity $e$ is given by:

If the tangent at a point on the ellipse $\frac{x^2}{27} + \frac{y^2}{3} = 1$ meets the coordinate axes at $A$ and $B,$ and $O$ is the origin,then the minimum area (in sq. units) of the triangle $OAB$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module