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Q.
The locus of the foot of the perpendicular from the centre of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ on a variable tan- gent is
Conic Sections
Solution:
Equation of any tangent to
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 ...$(1)
in the slope form is $y=m x+\sqrt{a^{2} m^{2}-b^{2}} ...$(2)
slope of tangent $=m$.
$\therefore $ slope of any line $\perp$ to it $=-\frac{1}{m}$
Equation of $\perp$ from centre $(0,0)$ of (1) on ( 2 ) is
$y-0=-\frac{1}{m}(x-0)$ or $ m=-\frac{x}{y}\,\,\,...$(3)
The required locus is obtained by eliminating the parameter $m$ between (2) and (3). Substituting for $m$ from (3) in (2), we get
$y=-\frac{x^{2}}{y}+\sqrt{a^{2} \cdot \frac{x^{2}}{y^{2}}-b^{2}}$
or $x^{2}+y^{2}=\sqrt{a^{2} x^{2}-b^{2} y^{2}}$
or $\left(x^{2}+y^{2}\right)^{2}=a^{2} x^{2}-b^{2} y^{2}$