Question

A variable line $\frac{x}{a}+\frac{y}{b}=1$ moves in such a manner so that the length of the perpendicular from origin on this line is constant and equal to $p$. If the line meets $x$-axis and $y$-axis at $A$ and $B$ respectively then the locus of the point of intersection of lines through $A$ and $B$ and perpendicular to axes is

• A. $x^2+y^2=p^2$
• B. $x+y=p$
• C. $x^2+y^2=p^2{x}^2{y}^2$
• D. $p^2(x^2+y^2)=x^2{y}^2$

Answer 1

Answer: (D)

$p^2(x^2+y^2)=x^2{y}^2$

Given that the length of the perpendicular from origin $=P$

It can be said that the line tangent if $x$-axis and $y$-axis meet at A and B respectively, then

$\Rightarrow$$x^2+y^2=P^2$

$\Rightarrow$Let $(x,y)$ $=(p\cos t,p\sin t)$ be the point of tangency.Its intercepts on $x$ and $y$ axes are $\frac{p}{\cos t}$ and $\frac{p}{\sin t}$

$\therefore$ $x=\frac{p}{\cos t}$ and $y=\frac{p}{\sin t}$

$\therefore \cos t=\frac{p}{x}$ and $\sin t=\frac{p}{y}$

Squaring and adding we get,

$\Rightarrow$${\cos }^{2}t+{\sin }^{2}t={\left(\frac{p}{x}\right)}^2+{\left(\frac{p}{y}\right)}^2$

$\Rightarrow$$\frac{p^2}{x^2}+\frac{p^2}{y^2}=1$

$\Rightarrow$$p^2(x^2+y^2)=x^2{y}^2$