Q) a variable line makes intercepts on coordinate axes . If the length of the perpendicular on the line from the origin is the geometric mean of the length of intercepts then find the locus of foot of perpendicular drawn from origin ?
My Attempt
Given that distance $p^{2}=ab$ And C is a point $(h,k)$ therefore we have $h^2 +k^2=ab$ . I also know that $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2} \Rightarrow a^2+b^2=ab$ .
Now how do i proceed ...?
On
Let the general equation of a line (L1) be of the form x/a+y/b=1 Let (h,k) lie on this line such that a normal line(L2) passing through this point intersects at the origin. Now we know that L1 is perpendicular to L2. Thus we get,
(k/h)*(-b/a)=-1
Which implies that:
k/h=a/b
Also, we know from the information provided in the question that ab=a^2+b^2, 1=a/b+b/a
From the manipulation we did above we get,
1=k/h+h/k which implies that
h^2+k^2=hk or x^2+y^2=xy.
Assume that $P=(x_0,y_0)$ is the projection of $O$ on the line.
The perpendicular to $OP$ through $P$ has equation:
$$ x_0 x + y y_0 = (x_0^2+y_0^2) $$ hence the lengths of the intercepts are given by $\frac{x_0^2+y_0^2}{x_0}$ and $\frac{x_0^2+y_0^2}{y_0}$.
The problem gives that $OP^2$ equals the product of these lengths, hence $x_0^2+y_0^2=x_0 y_0$ and $P$ belongs to the curve with equation $x^2-xy+y^2=0$, that is an ellipse.