Let \( F_1, F_2 \) \(\text{ be the foci of the hyperbola}\) \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, a > 0, \, b > 0, \] and let \( O \) be the origin. Let \( M \) be an arbitrary point on curve \( C \) and above the X-axis and \( H \) be a point on \( MF_1 \) such that \( MF_2 \perp F_1 F_2, \, M F_1 \perp OH, \, |OH| = \lambda |O F_2| \) with \( \lambda \in (2/5, 3/5) \), then the range of the eccentricity \( e \) is in:
Step 1: Understanding the eccentricity of a hyperbola.
The eccentricity \( e \) of a hyperbola is given by:
\[
e = \sqrt{1 + \frac{b^2}{a^2}}
\]
Step 2: The relationship between distances.
We are given the conditions involving the distances from \( M \) and \( O \) to the foci. Using the relationship between the distances and the specific bounds for \( \lambda \), we can derive the bounds for \( e \).
Step 3: Conclusion.
After applying the conditions and solving the range for \( e \), we find that the range of the eccentricity is \( (\sqrt{7/3}, 2) \), corresponding to option (b).
Thus, the correct answer is option (b).
Let \( C_{t-1} = 28, C_t = 56 \) and \( C_{t+1} = 70 \). Let \( A(4 \cos t, 4 \sin t), B(2 \sin t, -2 \cos t) \text{ and } C(3r - n_1, r^2 - n - 1) \) be the vertices of a triangle ABC, where \( t \) is a parameter. If \( (3x - 1)^2 + (3y)^2 = \alpha \) is the locus of the centroid of triangle ABC, then \( \alpha \) equals:
Consider the lines $ x(3\lambda + 1) + y(7\lambda + 2) = 17\lambda + 5 $. If P is the point through which all these lines pass and the distance of L from the point $ Q(3, 6) $ is \( d \), then the distance of L from the point \( (3, 6) \) is \( d \), then the value of \( d^2 \) is
"In order to be a teacher, one must graduate from college. All poets are poor. Some Mathematicians are poets. No college graduate is poor."
Which of the following is true?