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(C) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $1/a$ respectively,where $a 
eq 0$. The equation of the line in intercept

Vedclass · Accepted Answer

$4xy < 1$

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(C) Let the intercepts of the line on the $x$-axis and $y$-axis be $a$ and $1/a$ respectively,where $a 
eq 0$. The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{1/a} = 1$,which simplifies to $\frac{x}{a} + ay = 1$. Multiplying by $a$,we get $x + a^2y = a$,or $a^2y - a + x = 0$. Since $a$ is a real number,for the quadratic equation in $a$ to have real roots,the discriminant $D$ must be greater than or equal to $0$. Here,$D = (-1)^2 - 4(y)(x) \geq 0$. $1 - 4xy \geq 0 \Rightarrow 4xy \leq 1$. For a variable line,the point $P(x, y)$ on the line must satisfy $4xy < 1$ (excluding the boundary case where the line is fixed).

If a variable line is moving such that the intercepts made by it on the coordinate axes are reciprocal to each other,then the points $P(x, y)$ on such lines satisfy

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