(D) Given equations of straight lines are:$\frac{x}{a} + \frac{y}{b} = K$ $(1)$$\frac{x}{a} - \frac{y}{b} = \frac{1}{K}$ $(2)$Let the point of inter

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(D) Given equations of straight lines are:$\frac{x}{a} + \frac{y}{b} = K$ $(1)$$\frac{x}{a} - \frac{y}{b} = \frac{1}{K}$ $(2)$Let the point of intersection be $(x, y)$.Multiplying equation $(1)$ and $(2)$,we get:$(\frac{x}{a} + \frac{y}{b})(\frac{x}{a} - \frac{y}{b}) = K \times \frac{1}{K}$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$This is the equation of a hyperbola. Therefore,the locus is a hyperbola.

Class 11 Mathematics Straight Line Locus of Point

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The locus of the point of intersection of the straight lines $\frac{x}{a} + \frac{y}{b} = K$ and $\frac{x}{a} - \frac{y}{b} = \frac{1}{K}$,where $K$ is a non-zero real variable,is given by

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a straight line
B
an ellipse
C
a parabola
D
a hyperbola

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Straight Line

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Locus of Point

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Let $A(-1,1)$ and $B(2,3)$ be two points and $P(x,y)$ be a variable point above the line $AB$ such that the area of $\triangle PAB$ is $10$. If the locus of $P$ is $ax+by=15$,then $5a+2b$ is:

DifficultJEE MAIN 2024

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If the ends of the hypotenuse of a right-angled triangle are $(0, a)$ and $(a, 0)$,then the locus of the third vertex is:

EasyAP EAMCET 2023

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Suppose a point $P$ moves so that $BP^2 - AP^2 = 121$,where $A$ and $B$ are $(2, 5)$ and $(5, 11)$ respectively. Then the locus of $P$ is a straight line,whose slope is

MediumAP EAMCET 2022

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$A(a,0)$ and $B(-a,0)$ are two fixed points of triangle $ABC$. The vertex $C$ moves in such a way that $\cot A + \cot B = \lambda$,where $\lambda$ is a constant. Then the locus of the point $C$ is

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If the equation to the locus of points equidistant from the points $(-2, 3)$ and $(6, -5)$ is $a x + b y + c = 0$,where $a > 0$,then the ascending order of $a, b, c$ is

EasyAP EAMCET 2015

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The locus of the point of intersection of the straight lines $\frac{x}{a} + \frac{y}{b} = K$ and $\frac{x}{a} - \frac{y}{b} = \frac{1}{K}$,where $K$ is a non-zero real variable,is given by

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Let $A(-1,1)$ and $B(2,3)$ be two points and $P(x,y)$ be a variable point above the line $AB$ such that the area of $\triangle PAB$ is $10$. If the locus of $P$ is $ax+by=15$,then $5a+2b$ is:

If the ends of the hypotenuse of a right-angled triangle are $(0, a)$ and $(a, 0)$,then the locus of the third vertex is:

Suppose a point $P$ moves so that $BP^2 - AP^2 = 121$,where $A$ and $B$ are $(2, 5)$ and $(5, 11)$ respectively. Then the locus of $P$ is a straight line,whose slope is

$A(a,0)$ and $B(-a,0)$ are two fixed points of triangle $ABC$. The vertex $C$ moves in such a way that $\cot A + \cot B = \lambda$,where $\lambda$ is a constant. Then the locus of the point $C$ is

If the equation to the locus of points equidistant from the points $(-2, 3)$ and $(6, -5)$ is $a x + b y + c = 0$,where $a > 0$,then the ascending order of $a, b, c$ is

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