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
- 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 ______.
- Let the angles made with the positive $x$-axis by two straight lines drawn from the point $P(2, 3)$ and meeting the line $x + y = 6$ at a distance $\sqrt{\frac{2}{3}}$ from the point $P$ be $\theta_1$ and $\theta_2$. Then the value of $(\theta_1 + \theta_2)$ is:
- The sum of all values of $\alpha$, for which the shortest distance between the lines $\frac{x + 1}{\alpha} = \frac{y - 2}{-1} = \frac{z - 4}{-\alpha}$ and $\frac{x}{\alpha} = \frac{y - 1}{2} = \frac{z - 1}{2\alpha}$ is $\sqrt{2}$, is
- Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b)$, be 30. If its eccentricity is the maximum value of the function $f(t) = -\frac{3}{4} + 2t - t^2$, then $(a^2 + b^2)$ is equal to
- Let $\vec{a} = 2\hat{i} - \hat{j} - \hat{k}$, $\vec{b} = \hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} + \hat{j} + 3\hat{k}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to
- Let $\vec{a} = 2\hat{i} - 5\hat{j} + 5\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 3\hat{k}$. If $\vec{c}$ is a vector such that $2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0}$ and $(\vec{a} - \vec{b}) \cdot \vec{c} = -97$, then $|\vec{c} \times \hat{k}|^2$ is equal to
- Let the image of parabola $x^2 = 4y$, in the line $x - y = 1$ be $(y + a)^2 = b(x - c)$, $a, b, c \in \mathbb{N}$. Then $a + b + c$ is equal to
- Let $(h, k)$ lie on the circle $C : x^2 + y^2 = 4$ and the point $(2h + 1, 3k + 2)$ lie on an ellipse with eccentricity $e$. Then the value of $\frac{5}{e^2}$ is equal to _______.
- If $P$ is a point on the circle $x^2 + y^2 = 4$, $Q$ is a point on the straight line $5x + y + 2 = 0$ and $x - y + 1 = 0$ is the perpendicular bisector of $PQ$, then $13$ times the sum of abscissa of all such points $P$ is __________.
- If the distance of the point $\text{P}(43, \alpha, \beta)$, $\beta < 0$, from the line $\overrightarrow{\text{r}} = 4\hat{i} - \hat{k} + \mu(2\hat{i} + 3\hat{k})$, $\mu \in \mathbf{R}$ along a line with direction ratios $3, -1, 0$ is $13\sqrt{10}$, then $\alpha^{2} + \beta^{2}$ is equal to _________.
Important Questions from Coordinate Geometry
- 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 ______.
- Let the angles made with the positive $x$-axis by two straight lines drawn from the point $P(2, 3)$ and meeting the line $x + y = 6$ at a distance $\sqrt{\frac{2}{3}}$ from the point $P$ be $\theta_1$ and $\theta_2$. Then the value of $(\theta_1 + \theta_2)$ is:
- The sum of all values of $\alpha$, for which the shortest distance between the lines $\frac{x + 1}{\alpha} = \frac{y - 2}{-1} = \frac{z - 4}{-\alpha}$ and $\frac{x}{\alpha} = \frac{y - 1}{2} = \frac{z - 1}{2\alpha}$ is $\sqrt{2}$, is
- Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b)$, be 30. If its eccentricity is the maximum value of the function $f(t) = -\frac{3}{4} + 2t - t^2$, then $(a^2 + b^2)$ is equal to
- Let $\vec{a} = 2\hat{i} - \hat{j} - \hat{k}$, $\vec{b} = \hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} + \hat{j} + 3\hat{k}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to
