Let the product of the focal distances of the point $P(4,2\sqrt{3})$ on the hyperbola H: $\frac{x^2}{a^2} - \frac{y^2}{b^2}=1$ be 32. Let the length of the conjugate axis of H be $p$ and the length of its latus rectum be $q$. Then $p^2 + q^2$ is equal to
Hyperbola Parameters from Focal Distances
We are given a point $P(4, 2\sqrt{3})$ on the hyperbola H: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The product of the focal distances of P is 32.
1. Equation of Hyperbola
Since $P(4, 2\sqrt{3})$ lies on the hyperbola, its coordinates satisfy the equation:
$ \frac{4^2}{a^2} - \frac{(2\sqrt{3})^2}{b^2} = 1 $ $ \frac{16}{a^2} - \frac{12}{b^2} = 1 \quad (*) $2. Product of Focal Distances
For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the foci are at $(\pm c, 0)$, where $c^2 = a^2 + b^2$. The eccentricity is $e = c/a$. The focal distances of a point $(x, y)$ are $|ex \pm a|$.
The product of the focal distances is $|(ex+a)(ex-a)| = |e^2x^2 - a^2|$.
Given the product is 32 and the point is $P(4, 2\sqrt{3})$:
$ |e^2(4^2) - a^2| = 32 $ $ |16e^2 - a^2| = 32 $Substitute $e^2 = \frac{c^2}{a^2} = \frac{a^2+b^2}{a^2}$:
$ \left| 16 \left( \frac{a^2+b^2}{a^2} \right) - a^2 \right| = 32 $ $ \left| \frac{16(a^2+b^2) - a^4}{a^2} \right| = 32 $We consider two cases:
- Case 1: $\frac{16(a^2+b^2) - a^4}{a^2} = 32$ $ 16a^2 + 16b^2 - a^4 = 32a^2 $ $ 16b^2 = a^4 + 16a^2 $ From (*), $16b^2 - 12a^2 = a^2b^2$. Substitute $16b^2$: $ (a^4 + 16a^2) - 12a^2 = a^2b^2 $ $ a^4 + 4a^2 = a^2b^2 $ Since $a^2 \neq 0$, we get $a^2 + 4 = b^2$. Substitute $b^2 = a^2+4$ into $16b^2 = a^4 + 16a^2$: $ 16(a^2+4) = a^4 + 16a^2 $ $ 16a^2 + 64 = a^4 + 16a^2 $ $ a^4 = 64 \implies a^2 = 8 $ Then $b^2 = a^2 + 4 = 8 + 4 = 12$. Check with (*): $\frac{16}{8} - \frac{12}{12} = 2 - 1 = 1$. This case is valid.
- Case 2: $\frac{16(a^2+b^2) - a^4}{a^2} = -32$ $ 16a^2 + 16b^2 - a^4 = -32a^2 $ $ 16b^2 = a^4 - 48a^2 $ Substitute into $16b^2 - 12a^2 = a^2b^2$: $ (a^4 - 48a^2) - 12a^2 = a^2b^2 $ $ a^4 - 60a^2 = a^2b^2 $ $ a^2 - 60 = b^2 $ Since $b^2 > 0$, we must have $a^2 > 60$. Substitute $b^2$ into $16b^2 = a^4 - 48a^2$: $ 16(a^2 - 60) = a^4 - 48a^2 $ $ 16a^2 - 960 = a^4 - 48a^2 $ $ a^4 - 64a^2 + 960 = 0 $ Solving the quadratic for $a^2$ yields $a^2 = 40$ or $a^2 = 24$. In both sub-cases, $b^2 = a^2 - 60$ would be negative, which is impossible. Thus, Case 2 yields no valid solution.
We conclude that $a^2 = 8$ and $b^2 = 12$.
Calculating Conjugate Axis and Latus Rectum
The length of the conjugate axis is $p = 2b$. The length of the latus rectum is $q = \frac{2b^2}{a}$.
- From $b^2 = 12$, we get $b = \sqrt{12} = 2\sqrt{3}$. So, $p = 2b = 2(2\sqrt{3}) = 4\sqrt{3}$.
- From $a^2 = 8$, we get $a = \sqrt{8} = 2\sqrt{2}$. So, $q = \frac{2b^2}{a} = \frac{2(12)}{2\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2}$.
Final Calculation of $p^2 + q^2$
We need to find $p^2 + q^2$:
- $p^2 = (4\sqrt{3})^2 = 16 \times 3 = 48$.
- $q^2 = (6\sqrt{2})^2 = 36 \times 2 = 72$.
- $p^2 + q^2 = 48 + 72 = 120$.
Therefore, $p^2 + q^2 = 120$.
Similar Questions
The sum of all rational terms in the expansion of $(2+\sqrt{3})^8$ is
The number of solutions of the equation $2x + 3\tan x = \pi$, $x \in [-2\pi, 2\pi]-\left\{ \pm \frac{\pi}{2}, \pm \frac{3\pi}{2} \right\}$ is:
A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines $L_1: 2x + y + 6 = 0$ and $L_2: 4x+2y-p = 0$, $p > 0$, at the points A and B, respectively. If $AB = \frac{9}{\sqrt{2}}$ and the foot of the perpendicular from the point A on the line $L_2$ is M, then $\frac{AM}{BM}$ is equal to
Line $L_1$ passes through the point $(1, 2, 3)$ and is parallel to z-axis. Line $L_2$ passes through the point $(\lambda, 5, 6)$ and is parallel to y-axis. Let for $\lambda = \lambda_1, \lambda_2, \lambda_2 < \lambda_1$, the shortest distance between the two lines be 3. Then the square of the distance of the point $(\lambda_1, \lambda_2, 7)$ from the line $L_1$ is
Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=3\hat{i}+2\hat{j}-\hat{k}$, $\vec{c} = \lambda\hat{j} + \mu\hat{k}$ and $\vec{d}$ be a unit vector such that $\vec{a}\times\vec{d}=\vec{b}\times\vec{d}$ and $\vec{c}\cdot\vec{d} = 1$. If $\vec{c}$ is perpendicular to $\vec{a}$, then $|3 \lambda\vec{d} + \mu\vec{c}|^2$ is equal to
Let the focal chord PQ of the parabola $y^2=4x$ make an angle of $60^\circ$ with the positive x-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, S being the focus of the parabola, touches the y-axis at the point (0, $\alpha$), then $5\alpha^2$ is equal to :
If S and S' are the foci of the ellipse $\frac{x^2}{18} + \frac{y^2}{9} = 1$ and P be a point on the ellipse, then min $(SP \cdot S'P)$ + max $(SP \cdot S'P)$ is equal to :
- Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the $\triangle BCD$ is equal to :
- Let one focus of the hyperbola H: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be at $(\sqrt{10},0)$ and the corresponding directrix be $x = \frac{9}{\sqrt{10}}$. If e and l respectively are the eccentricity and the length of the latus rectum of H, then $9(e^2 + l)$ is equal to :
- If $\vec{a}$ is a nonzero vector such that its projections on the vectors $2\hat{i}-\hat{j}+2\hat{k}$, $\hat{i}+2\hat{j}-2\hat{k}$ and $\hat{k}$ are equal, then a unit vector along $\vec{a}$ is:
Important Questions from Coordinate Geometry
The sum of all rational terms in the expansion of $(2+\sqrt{3})^8$ is
The number of solutions of the equation $2x + 3\tan x = \pi$, $x \in [-2\pi, 2\pi]-\left\{ \pm \frac{\pi}{2}, \pm \frac{3\pi}{2} \right\}$ is:
A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines $L_1: 2x + y + 6 = 0$ and $L_2: 4x+2y-p = 0$, $p > 0$, at the points A and B, respectively. If $AB = \frac{9}{\sqrt{2}}$ and the foot of the perpendicular from the point A on the line $L_2$ is M, then $\frac{AM}{BM}$ is equal to
Line $L_1$ passes through the point $(1, 2, 3)$ and is parallel to z-axis. Line $L_2$ passes through the point $(\lambda, 5, 6)$ and is parallel to y-axis. Let for $\lambda = \lambda_1, \lambda_2, \lambda_2 < \lambda_1$, the shortest distance between the two lines be 3. Then the square of the distance of the point $(\lambda_1, \lambda_2, 7)$ from the line $L_1$ is
Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=3\hat{i}+2\hat{j}-\hat{k}$, $\vec{c} = \lambda\hat{j} + \mu\hat{k}$ and $\vec{d}$ be a unit vector such that $\vec{a}\times\vec{d}=\vec{b}\times\vec{d}$ and $\vec{c}\cdot\vec{d} = 1$. If $\vec{c}$ is perpendicular to $\vec{a}$, then $|3 \lambda\vec{d} + \mu\vec{c}|^2$ is equal to
