Solving for Chord Property on Parabola
We are given a parabola defined by the equation $y^2 = 12x$ and two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ on it. The chord connecting these points, $P_1P_2$, subtends a right angle at the vertex of the parabola, $V(0,0)$. We need to find the value of the expression $x_1x_2 - y_1y_2$.
Parabola Analysis
The standard equation of a parabola with vertex at the origin is $y^2 = 4ax$. Comparing this with the given equation $y^2 = 12x$, we find $4a = 12$, which implies $a = 3$. The vertex $V$ is at $(0,0)$.
Using Parametric Coordinates
Points on the parabola $y^2 = 4ax$ can be represented parametrically as $(at^2, 2at)$. For our parabola $y^2 = 12x$ (where $a=3$), the points are of the form $(3t^2, 6t)$.
- Let $P_1$ correspond to parameter $t_1$, so $P_1 = (3t_1^2, 6t_1)$. This means $x_1 = 3t_1^2$ and $y_1 = 6t_1$.
- Let $P_2$ correspond to parameter $t_2$, so $P_2 = (3t_2^2, 6t_2)$. This means $x_2 = 3t_2^2$ and $y_2 = 6t_2$.
Condition for Right Angle at Vertex
The chord $P_1P_2$ subtending a right angle at the vertex $V(0,0)$ means the line segments $VP_1$ and $VP_2$ are perpendicular.
The slope of $VP_1$ is $m_1 = \frac{y_1 - 0}{x_1 - 0} = \frac{6t_1}{3t_1^2} = \frac{2}{t_1}$ (assuming $t_1 \neq 0$).
The slope of $VP_2$ is $m_2 = \frac{y_2 - 0}{x_2 - 0} = \frac{6t_2}{3t_2^2} = \frac{2}{t_2}$ (assuming $t_2 \neq 0$).
The condition for perpendicularity is $m_1 m_2 = -1$. $ \implies \left(\frac{2}{t_1}\right) \left(\frac{2}{t_2}\right) = -1 $ $ \implies \frac{4}{t_1 t_2} = -1 $ $ \implies t_1 t_2 = -4 $
Calculating the Expression
We need to evaluate $x_1x_2 - y_1y_2$. Using the parametric forms:
- $x_1x_2 = (3t_1^2)(3t_2^2) = 9 (t_1 t_2)^2$
- $y_1y_2 = (6t_1)(6t_2) = 36 (t_1 t_2)$
Substitute the derived value $t_1 t_2 = -4$:
- $x_1x_2 = 9 (-4)^2 = 9(16) = 144$
- $y_1y_2 = 36 (-4) = -144$
Now, compute the final expression:
$ x_1x_2 - y_1y_2 = 144 - (-144) = 144 + 144 = 288 $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 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
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 :
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 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
