Question

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

• A. ${\left(x^2+y^2\right)}^2=a^2{x}^2+b^2{y}^2$
• B. ${\left(x^2+y^2\right)}^2=a^2{x}^2-b^2{y}^2$
• C. ${\left(x^2+y^2\right)}^2=b^2{x}^2-a^2{y}^2$
• D. none of these

Answer 1

Answer: (B)

Equation of any tangent to
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=\mathrm{1...}$(1)
in the slope form is $y=mx+\sqrt{a^2{m}^2-b^2}...$(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 $m=-\frac{x}{y}$
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 ${\left(x^2+y^2\right)}^2=a^2{x}^2-b^2{y}^2$