Question

The locus of foot of perpendicular from any focus of a hyperbola upon any tangent to the hyperbola is the auxiliary circle of the hyperbola. Consider the foci of a hyperbola as $(-3,-2)$ and $(5,6)$ and the foot of perpendicular from the focus $(5,6)$ upon tangent to the hyperbola as $(2,5).$ Then

• A. radius of the auxiliary circle is $\sqrt{10}$
• B. conjugate axis of the hyperbola is $2\sqrt{22}$
• C. the directrix of the hyperbola corresponding to the focus $(5,6)$ is $2x+2y=11$
• D. the point of contact of the tangent with the hyperbola is $(\frac{2}{3},9)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A radius of the auxiliary circle is $\sqrt{10}$
B conjugate axis of the hyperbola is $2\sqrt{22}$
C the directrix of the hyperbola corresponding to the focus $(5,6)$ is $2x+2y=11$
D the point of contact of the tangent with the hyperbola is $(\frac{2}{3},9)$

Centre $\equiv (1,2)$
Radius of auxiliary circle is given by,
$a=\sqrt{(2-1{)}^2+(5-2{)}^2}=\sqrt{10}$

$2ae=\sqrt{8^2+8^2}=8\sqrt{2}$
$\Rightarrow e=\frac{4}{\sqrt{5}}$
$\Rightarrow b^2=a^2{e}^2-a^2=32-10=22$
$\therefore 2b=2\sqrt{22}$
Hence, conjugate axis of the hyperbola is $2\sqrt{22}.$

Directrix is perpendicular to the transverse axis.
Let the equation of directrix be $x+y=k$
Its distance from the centre is,
$\frac{a}{e}=\frac{|1+2-k|}{\sqrt{2}}$
$\Rightarrow \pm \frac{5}{2\sqrt{2}}=\frac{3-k}{\sqrt{2}}$
$\Rightarrow k=3\pm \frac{5}{2}=\frac{1}{2}\text{or}\frac{11}{2}$
Therefore, the directrix of the hyperbola corresponding to the focus $(5,6)$ is $2x+2y=11$

Equation of tangent is,
$y-5=-\frac{5-2}{6-5}(x-2)$
$\Rightarrow 3x+y=11$

The hyperbola is,
$(x-5{)}^2+(y-6{)}^2=\frac{16}{5}\times \frac{(2x+2y-11{)}^2}{8}$
$\Rightarrow (x-5{)}^2+(5-3x{)}^2=\frac{2}{5}\times (2x+22-6x-11{)}^2$
$\Rightarrow 9x^2-12x+4=0\Rightarrow (3x-2{)}^2=0$
$\Rightarrow x=\frac{2}{3},y=9$
Therefore, the point of contact of the tangent with the hyperbola is $(\frac{2}{3},9)$