Ellipse and Hyperbola Foci Concordance Calculation
The problem requires finding the value of \( 24(e + L) \) where \( e \) is the eccentricity and \( L \) is the length of the latus rectum of a hyperbola \( \text{H} \), given that \( \text{H} \) shares the same foci as a given ellipse \( \text{E} \).
Foci of Ellipse E
The equation of the ellipse \( \text{E} \) is given by:
$ \frac{x^{2}}{144} + \frac{y^{2}}{169} = 1 $
This is of the form \( \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 \) since \( 169 > 144 \). Here, \( a^{2} = 169 \) and \( b^{2} = 144 \). The major axis is along the y-axis.
The distance from the center to the foci is given by \( c \), where \( c^{2} = a^{2} - b^{2} \).
$ c^{2} = 169 - 144 = 25 $
$ c = \sqrt{25} = 5 $
Since the major axis is along the y-axis, the foci of the ellipse \( \text{E} \) are at \( (0, \pm c) \), which are \( (0, \pm 5) \).
Hyperbola H Parameters
The equation of the hyperbola \( \text{H} \) is:
$ \frac{x^{2}}{16} - \frac{y^{2}}{\lambda^{2}} = -1 $
This can be rewritten in standard form as:
$ \frac{y^{2}}{\lambda^{2}} - \frac{x^{2}}{16} = 1 $
This represents a vertical hyperbola of the form \( \frac{y^{2}}{A^{2}} - \frac{x^{2}}{B^{2}} = 1 \). Comparing the equations, we have \( A^{2} = \lambda^{2} \) and \( B^{2} = 16 \).
The foci of this vertical hyperbola are at \( (0, \pm C) \), where \( C^{2} = A^{2} + B^{2} \).
Since the hyperbola \( \text{H} \) has the same foci as the ellipse \( \text{E} \), we must have \( C = c = 5 \).
$ C^{2} = A^{2} + B^{2} $
$ 5^{2} = \lambda^{2} + 16 $
$ 25 = \lambda^{2} + 16 $
$ \lambda^{2} = 25 - 16 = 9 $
Thus, for the hyperbola \( \text{H} \), we have \( A^{2} = 9 \) and \( B^{2} = 16 \). This implies \( A = \sqrt{9} = 3 \) and \( B = \sqrt{16} = 4 \).
Eccentricity (e) of H
The eccentricity \( e \) of a vertical hyperbola \( \frac{y^{2}}{A^{2}} - \frac{x^{2}}{B^{2}} = 1 \) is given by the formula:
$ e = \sqrt{1 + \frac{B^{2}}{A^{2}}} $
Substituting the values \( A^{2} = 9 \) and \( B^{2} = 16 \):
$ e = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{9 + 16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} $
Latus Rectum (L) of H
The length of the latus rectum \( L \) for a vertical hyperbola is given by the formula:
$ L = \frac{2B^{2}}{A} $
Substituting the values \( A = 3 \) and \( B^{2} = 16 \):
$ L = \frac{2 \times 16}{3} = \frac{32}{3} $
Final Calculation
We need to calculate the value of \( 24(e + L) \).
First, find the sum \( e + L \):
$ e + L = \frac{5}{3} + \frac{32}{3} = \frac{5 + 32}{3} = \frac{37}{3} $
Now, multiply this sum by 24:
$ 24(e + L) = 24 \times \frac{37}{3} $
$ 24(e + L) = \frac{24}{3} \times 37 = 8 \times 37 $
$ 8 \times 37 = 296 $
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