JEE 2023 Mathematics PYQ — An arc of a circle subtends a right angle at its centre . The mid… | Mathem Solvex | Mathem Solvex
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JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023
An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. OP=u, OR=v and OQ=αu+βv, then α,β2 are the roots of the equation:
Choose the correct answer:
A.
3x2−2x−1=0
B.
3x2+2x−1=0
C.
x2−x−2=0
(Correct Answer)
D.
x2+x−2=0
Correct Answer:
x2−x−2=0
Explanation
1. Identify the Geometry
Let the radius of the circle be r. Since P,Q, and R lie on the circle:
∣u∣=∣v∣=∣OQ∣=r
The angle between OP (u) and OQ is 90∘ (right angle).
Since R is the midpoint of arc PQ, the angle between u and v is 45∘, and the angle between v and OQ is also 45∘.
2. Set up the Dot Products
Using the dot product formula a⋅b=∣a∣∣b∣cosθ:
u⋅OQ=r⋅rcos90∘=0
u⋅v=r⋅rcos45∘=2r2
v⋅OQ=r⋅rcos45∘=2r2
3. Use the Vector Equation
We are given:
OQ=αu+βv
Dot product with u:
u⋅OQ=α(u⋅u)+β(u⋅v)
0=αr2+β(2r2)
Divide by r2:
0=α+2β⟹β=−2α…(Eq. 1)
Dot product with v:
v⋅OQ=α(v⋅u)+β(v⋅v)
2r2=α(2r2)+βr2
Divide by r2:
21=2α+β…(Eq. 2)
4. Solve for α and β2
Substitute β=−2α into Eq. 2:
21=2α−2α
Multiply by 2:
1=α−2α⟹1=−α⟹α=−1
Now find β:
β=−2(−1)=2
So, β2=(2)2=2.
5. Form the Quadratic Equation
The roots are α=−1 and β2=2.
The equation is:
(x−α)(x−β2)=0
(x−(−1))(x−2)=0
(x+1)(x−2)=0
x2−2x+x−2=0
x2−x−2=0
Explanation
1. Identify the Geometry
Let the radius of the circle be r. Since P,Q, and R lie on the circle:
∣u∣=∣v∣=∣OQ∣=r
The angle between OP (u) and OQ is 90∘ (right angle).
Since R is the midpoint of arc PQ, the angle between u and v is 45∘, and the angle between v and OQ is also 45∘.
