JEE 2023 Mathematics PYQ — The area enclosed by the closed curve given by the differential e… | Mathem Solvex | Mathem Solvex
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JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023
The area enclosed by the closed curve C given by the differential equation dxdy+y−2x+a=0,y(1)=0 is 4π. Let P and Q be the points of intersection of the curve C and the y-axis. If normals at P and Q on the curve C intersect x-axis at points R and S respectively, then the length of the line segment RS is:
Choose the correct answer:
A.
2
B.
343
(Correct Answer)
C.
23
D.
323
Correct Answer:
343
Explanation
Solution
1. Solving the Differential Equation
The given equation is:
dxdy=−y−2x+a
By separating the variables, we get:
(y−2)dy=−(x+a)dx
Integrating both sides:
∫(y−2)dy=−∫(x+a)dx
2(y−2)2=−2(x+a)2+k
(x+a)2+(y−2)2=2k
Let 2k=r2. This is the equation of a circle centered at (−a,2) with radius r.
2. Finding the value of 'a' and 'r'
The problem states the area of the closed curve is 4π.
Area=πr2=4π⟹r2=4⟹r=2
The curve passes through (1,0) as y(1)=0:
(1+a)2+(0−2)2=4
(1+a)2+4=4
(1+a)2=0⟹a=−1
So, the equation of the circle C is:
(x−1)2+(y−2)2=4
3. Finding points P and Q
P and Q are points where the curve intersects the y-axis (where x=0):
(0−1)2+(y−2)2=4
1+(y−2)2=4⟹(y−2)2=3
y−2=±3⟹y=2±3
Thus, the points are P(0,2+3) and Q(0,2−3).
4. Finding points R and S
A key property of a circle is that the normal at any point passes through the center.
The center of our circle is M(1,2).
Normal at P: Passes through P(0,2+3) and M(1,2).
The slope mP=1−02−(2+3)=−3.
Equation: y−2=−3(x−1).
To find R (intersection with x-axis, y=0):
−2=−3(x−1)⟹x−1=32⟹xR=1+32.
Normal at Q: Passes through Q(0,2−3) and M(1,2).
The slope mQ=1−02−(2−3)=3.
Equation: y−2=3(x−1).
To find S (intersection with x-axis, y=0):
−2=3(x−1)⟹x−1=−32⟹xS=1−32.
5. Calculating Length RS
RS=∣xR−xS∣
RS=(1+32)−(1−32)
RS=34=343
Correct Option: (2)
Explanation
Solution
1. Solving the Differential Equation
The given equation is:
dxdy=−y−2x+a
By separating the variables, we get:
(y−2)dy=−(x+a)dx
Integrating both sides:
∫(y−2)dy=−∫(x+a)dx
2(y−2)2=−2(x+a)2+k
(x+a)2+(y−2)2=2k
Let 2k=r2. This is the equation of a circle centered at (−a,2) with radius r.
2. Finding the value of 'a' and 'r'
The problem states the area of the closed curve is 4π.
Area=πr2=4π⟹r2=4⟹r=2
The curve passes through (1,0) as y(1)=0:
(1+a)2+(0−2)2=4
(1+a)2+4=4
(1+a)2=0⟹a=−1
So, the equation of the circle C is:
(x−1)2+(y−2)2=4
3. Finding points P and Q
P and Q are points where the curve intersects the y-axis (where x=0):
(0−1)2+(y−2)2=4
1+(y−2)2=4⟹(y−2)2=3
y−2=±3⟹y=2±3
Thus, the points are P(0,2+3) and Q(0,2−3).
4. Finding points R and S
A key property of a circle is that the normal at any point passes through the center.
The center of our circle is M(1,2).
Normal at P: Passes through P(0,2+3) and M(1,2).
The slope mP=1−02−(2+3)=−3.
Equation: y−2=−3(x−1).
To find R (intersection with x-axis, y=0):
−2=−3(x−1)⟹x−1=32⟹xR=1+32.
Normal at Q: Passes through Q(0,2−3) and M(1,2).
The slope mQ=1−02−(2−3)=3.
Equation: y−2=3(x−1).
To find S (intersection with x-axis, y=0):
−2=3(x−1)⟹x−1=−32⟹xS=1−32.
5. Calculating Length RS
RS=∣xR−xS∣
RS=(1+32)−(1−32)
RS=34=343
Correct Option: (2)
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