A company situated at (2,0) in the XY-plane c | vedclass

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

Question

(B) Let $P(x, y)$ be any point in the $XY$-plane.According to the problem,the cost of the first company is $2 	imes PA$ and the second is $3 	imes P

Vedclass · Accepted Answer

the outside of the circle $(x+1.6)^2+(y-5.4)^2=18.72$

Vedclass · Answer

(B) Let $P(x, y)$ be any point in the $XY$-plane.According to the problem,the cost of the first company is $2 	imes PA$ and the second is $3 	imes PB$.We want the region where it is cheaper to use the first company,so $2 PA Squaring both sides,we get $4 PA^2 Substituting the coordinates $A(2, 0)$ and $B(0, 3)$,we have $4[(x-2)^2 + y^2] Expanding this,$4(x^2 - 4x + 4 + y^2) $4x^2 - 16x + 16 + 4y^2 Rearranging terms,$5x^2 + 5y^2 + 16x - 54y + 65 > 0$.Dividing by $5$,$x^2 + y^2 + 3.2x - 10.8y + 13 > 0$.Completing the square,$(x^2 + 3.2x + 1.6^2) + (y^2 - 10.8y + 5.4^2) > -13 + 2.56 + 29.16$.$(x + 1.6)^2 + (y - 5.4)^2 > 18.72$.This represents the region outside the circle $(x + 1.6)^2 + (y - 5.4)^2 = 18.72$.

$A$ company situated at $(2,0)$ in the $XY$-plane charges $RS. 2$ per $km$ for delivery. $A$ second company at $(0,3)$ charges $RS. 3$ per $km$ for delivery. The region of the plane where it is cheaper to use the first company is

Similar Questions

The locus of the third vertex of a right-angled triangle,the ends of whose hypotenuse are $(1,2)$ and $(4,5)$,is

$A$ point moves in such a way that the sum of the squares of its distances from the points $A(2, 0)$ and $B(-2, 0)$ is always equal to the square of the distance between $A$ and $B$. The locus of the point is

$A$ point moves in such a way that its distance from the origin is always $4$. Then the locus of the point is:

Two straight rods of lengths $2a$ and $2b$ move along the coordinate axes in such a way that their extremities are always concyclic. Then the locus of the centres of such circles is

Vedclass Test Series

Exam Paper Generator

Online Exam Module