Question

If \( l \) is the maximum value of \( -3x^2+4x+1 \) and \( m \) is the minimum value of \( 3x^2+4x+1 \), then the equation of the hyperbola having foci at \( (l,0), (7m,0) \) and eccentricity as 2 is

• A. \( 36x^2 - 12y^2 = 49 \)
• B. \( 2x^2 - 5y^2 = 1 \)
• C. \( 49x^2 - 36y^2 = 12 \)
• D. \( 36x^2 - 12y^2 = 1 \)

Hint:

For a quadratic \( ax^2 + bx + c \), the vertex y-coordinate (max/min value) is given directly by \( -\frac{D}{4a} \) where \( D = b^2 - 4ac \). This avoids substituting the x-value back. For \( -3x^2+4x+1 \): \( D = 16 - 4(-3)(1) = 28 \). Max = \( -28 / (-12) = 7/3 \). Correct.

Answer 1

The Correct Option is A

Step 1: Find \( l \) and \( m \):

1. Find \( l \): \( f(x) = -3x^2 + 4x + 1 \). This is a downward parabola. Maximum occurs at vertex \( x = \frac{-b}{2a} = \frac{-4}{2(-3)} = \frac{2}{3} \). \( l = f(2/3) = -3(\frac{4}{9}) + 4(\frac{2}{3}) + 1 = -\frac{4}{3} + \frac{8}{3} + 1 = \frac{4}{3} + 1 = \frac{7}{3} \). 2. Find \( m \): \( g(x) = 3x^2 + 4x + 1 \). This is an upward parabola. Minimum occurs at vertex \( x = \frac{-b}{2a} = \frac{-4}{2(3)} = -\frac{2}{3} \). \( m = g(-2/3) = 3(\frac{4}{9}) + 4(-\frac{2}{3}) + 1 = \frac{4}{3} - \frac{8}{3} + 1 = -\frac{4}{3} + 1 = -\frac{1}{3} \).
Step 2: Determine Foci and Parameters:

The foci are given as \( (l, 0) \) and \( (7m, 0) \). \( F_1 = (\frac{7}{3}, 0) \) \( F_2 = (7(-\frac{1}{3}), 0) = (-\frac{7}{3}, 0) \) Since the foci are symmetric about the origin on the x-axis, the center of the hyperbola is \( (0,0) \), and it is a horizontal hyperbola. The distance between foci is \( 2ae \). \[ 2ae = \frac{7}{3} - \left(-\frac{7}{3}\right) = \frac{14}{3} \] Given eccentricity \( e = 2 \): \[ 2a(2) = \frac{14}{3} \implies 4a = \frac{14}{3} \implies a = \frac{14}{12} = \frac{7}{6} \] \[ a^2 = \frac{49}{36} \] Now find \( b^2 \) using \( b^2 = a^2(e^2 - 1) \): \[ b^2 = \frac{49}{36} (2^2 - 1) = \frac{49}{36} (3) = \frac{49}{12} \]
Step 3: Equation of Hyperbola:

Equation: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) \[ \frac{x^2}{49/36} - \frac{y^2}{49/12} = 1 \] \[ \frac{36x^2}{49} - \frac{12y^2}{49} = 1 \] Multiply by 49: \[ 36x^2 - 12y^2 = 49 \]
Step 4: Final Answer:

The equation is \( 36x^2 - 12y^2 = 49 \).