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
- 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 :
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- 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 ______.
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- 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
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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 ______.
