Tag: auxiliary circle

Equation of tangent of given hyperbola at point
$\displaystyle (h,\, k)\,$ is $\dfrac{hx}{a^2}\, -\, \dfrac{ky}{b^2}\, =\, 1$ ...(i)
Equation of auxillary circle is $x^2\, +\, y^2\, =\, a^2$ .....(ii)
From (i) and (ii)
$\displaystyle \left [ \left ( 1\, +\, \frac{ky}{b^2}\right ) \frac{a^2}{h}\right ]^2\, +\, y^2\, -\, a^2\, =\, 0$
$\Rightarrow\, y^2\, (k^2a^4\, +\, b^4h^2)\, +\, 2kb^2a^4y\, +\, b^4a^2\, (a^2\, -\, h^2)\, =\, 0$
Now  $\displaystyle \, \frac{y _1\, +\, y _2}{y _1y _2}\, =\, -\,

\frac{2kb^2a^4}{b^4a^2(a^2\, -\, h^2)}\, =\, \frac{-2ka^2}{b^2a^2 \left (

1\, -\, \frac{h^2}{a^2}\right )}$
$\displaystyle =\, \frac{-2k}{b^2 \left ( \frac{-k^2}{b^2}\right )}\, =\, \frac{2}{k}$

The director circle is the locus of the point of intersection of a pair of perpendicular tangents to a hyperbola.

Equation of the director circle of the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ is $x^2 + y^2 = a^2  b^2$ i.e. a circle whose center is origin and radius is $(a^2  b^2)$.

Hence, for the director circle to be a point circle, $a^2 = b^2$ .

$p^2 - 4 = p^2 + 4p + 3$ ---> $4p = -7$ ---> $p = -\dfrac{7}{4}$ . Hence, option (B).