Question

A point `P` moves between lines `y=0` and `y=mx` so that the area of quadrilateral formed by the two lines and perpendicular from `P` on these lines remains constant. Find the equation of locus of `P`?

Answer 1

The equation is of the locus is of type `my^2-mx^2+2xy=k` and it is a hyperbola.

Explanation:

Let us consider that equation of line `OA` is `y=0` (`x`-axis) and that of line `OB` is `y=mx` and `O` is origin. Let the coordinates of `P` be `(x,y)`. The diagram appears as follows:

Point `P(x,y)` moves so that area of quarilateral `OMPN` is some constant.

Now it is evident that area of `DEltaPMO=1/2xy`. For area of `DeltaPNO`, and `OP=sqrt(x^2+y^2)`. Let us work out `PN` and `ON`. It is evident that `PN=|(y-mx)/sqrt(1+m^2)|`

and hence `ON^2=x^2+y^2-(y-mx)^2/(1+m^2)`

= `(x^2+y^2+m^2x^2+m^2y^2-y^2-m^2x^2+2mxy)/(1+m^2)`

= `(x^2+m^2y^2+2mxy)/(1+m^2)=(x+my)^2/(1+m^2)`

and `OP=|(x+my)/sqrt(1+m^2)|`

Hence area of `DeltaPNO` is

`1/2((y-mx)(x+my))/(1+m^2)=1/2(my^2-mx^2+xy-m^2xy)/(1+m^2)`

and area of quadrilateral is

`1/2xy+1/2(my^2-mx^2+xy-m^2xy)/(1+m^2)`

= `1/2((xy+xym^2+my^2-mx^2+xy-m^2xy)/(1+m^2))`

= `1/2((xy+my^2-mx^2+xy)/(1+m^2))`

Hence equation of `P(x,y)` would be

`my^2-mx^2+2xy=k`, where `k` is a constant.

This is the equation of a hyperbola.

Below is shown the graph for `m=2` and `k=20`

graph{y^2-x^2+xy=10 [-20, 20, -10, 10]}