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

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Chapter

Straight Line

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Locus of Point

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If $A=(5,3)$,$B=(3,-2)$ and a point $P$ is such that the area of the triangle $PAB$ is $9$,then the locus of $P$ represents

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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

Similar Questions

$A$ particle is projected vertically upwards. If it has to stay above the ground for $12 \text{ seconds}$, then:

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$A$ straight line passing through a point $(3, 2)$ cuts the $X$ and $Y$-axes at points $A$ and $B$ respectively. If a point $P(h, k)$ divides $AB$ in the ratio $2: 3$,then the equation of the locus of point $P$ is

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