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 $
Similar Questions
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Important Questions from Coordinate Geometry
- If the line $\alpha x + 2y = 1$, where $\alpha \in \mathbb{R}$, does not meet the hyperbola $x^2 - 9y^2 = 9$, then a possible value of $\alpha$ is:
- Let $P(10, 2\sqrt{15})$ be a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, whose foci are S and $S'$. If the length of its latus rectum is 8, then the square of the area of $\Delta PSS'$ is equal to :
- Let the locus of the mid-point of the chord through the origin O of the parabola $y^2 = 4x$ be the curve S. Let P be any point on S. Then the locus of the point, which internally divides OP in the ratio 3:1, is :
- Let a circle of radius 4 pass through the origin O, the points $A(-\sqrt{3}a, 0)$ and $B(0, -\sqrt{2}b)$, where $a$ and $b$ are real parameters and $ab \neq 0$. Then the locus of the centroid of $\Delta OAB$ is a circle of radius
- Let a line L passing through the point $P(1, 1, 1)$ be perpendicular to the lines $\frac{x-4}{4} = \frac{y-1}{1} = \frac{z-1}{1}$ and $\frac{x-17}{1} = \frac{y-71}{1} = \frac{z}{0}$. Let the line L intersect the yz-plane at the point Q. Another line parallel to L and passing through the point $S(1, 0, -1)$ intersects the yz-plane at the point R. Then the square of the area of the parallelogram PQRS is equal to ______.
