NIMCET 2025 Mathematics PYQ — Let be the foci of the hyperbola ( ), and let be the origin. Let … | Mathem Solvex | Mathem Solvex
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NIMCET 2025 — Mathematics PYQ
NIMCET | Mathematics | 2025
Let F1,F2 be the foci of the hyperbola a2x2−b2y2=1 (a>0, b>0), and let O be the origin. Let M be an arbitrary point on curve C above the X-axis such that MF2⊥F1F2. Let H be a point on MF1 such that MF1⊥OH. If ∣OH∣=λ∣OF2∣ with λ∈(52,53), then find the range of the eccentricity e.
Choose the correct answer:
A.
(1,7/3)
B.
(7/3,2)
(Correct Answer)
C.
(2,3)
D.
(3,2)
Correct Answer:
(7/3,2)
Explanation
1. Identify the Coordinates of the Points
For the standard hyperbola a2x2−b2y2=1, the foci lie on the X-axis. Let:
O=(0,0)
F1=(−ae,0)
F2=(ae,0)
Since MF2⊥F1F2 and M lies above the X-axis, the x-coordinate of M is ae. Substituting x=ae into the hyperbola equation:
a2(ae)2−b2y2=1⟹e2−1=b2y2
Since b2=a2(e2−1), we get y2=a2b4, which gives y=ab2 (as M is above the X-axis).
Thus, the coordinates of M are:
M=(ae,ab2)
2. Equation of the Line MF1
Let's find the equation of the line MF1 passing through F1(−ae,0) and M(ae,ab2):
Slope m=ae−(−ae)ab2−0=2a2eb2
Using b2=a2(e2−1), the slope becomes:
m=2a2ea2(e2−1)=2ee2−1
The equation of line MF1 is:
y−0=2ee2−1(x+ae)
(e2−1)x−2ey+ae(e2−1)=0
3. Calculate the Length ∣OH∣
The perpendicular distance ∣OH∣ from the origin O(0,0) to this line is given by:
∣OH∣=(e2−1)2+(2e)2∣(e2−1)(0)−2e(0)+ae(e2−1)∣
Simplify the denominator:
(e2−1)2+4e2=e4−2e2+1+4e2=e4+2e2+1=(e2+1)2
Thus, the length OH is:
∣OH∣=e2+1ae(e2−1)
4. Use the Given Condition to Find the Range of e
We are given that ∣OH∣=λ∣OF2∣. Since the distance ∣OF2∣=ae, we substitute the values:
