Question

The chords of contact of a point P w.r.t. a hyperbola and its auxiliary circle are at right angle , then the point P lies on
A.Conjugate hyperbola
B.One of the directrix
C.One of the asymptotes
D.None of these

Answer 1

Let the point P be (h,k) .The chord of contact of point P (h,k) with respect to the hyperbola is given by $\frac{{xh}}{{{a^2}}} - \frac{{yk}}{{{b^2}}} = 1$ and the chord of contact of a point $\left( {{x_1},{y_1}} \right)$ with respect to its auxiliary circle is given by $x{x_1} + y{y_1} = {a^2} - {b^2}$. Find the slope of the two equations and since we are given that they are at right angles we get the product of the slopes to be – 1 . And y further simplification we get the combined equation of the asymptotes of the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$.Hence the point lies on one of the asymptotes.

Complete step-by-step answer:
Step 1:
Let P (h,k) be any point.
We know that any chord of contact of point$\left( {{x_1},{y_1}} \right)$ with respect to the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$is given by T = 0 where T = $\frac{{x{x_1}}}{{{a^2}}} - \frac{{y{y_1}}}{{{b^2}}} - 1$
Therefore the chord of contact of point$\left( {{x_1},{y_1}} \right)$is given by $\frac{{x{x_1}}}{{{a^2}}} - \frac{{y{y_1}}}{{{b^2}}} = 1$
Now using the above formula , the chord of contact of point P (h,k) with respect to the hyperbola is given by $\frac{{xh}}{{{a^2}}} - \frac{{yk}}{{{b^2}}} = 1$………….(1)
Step 2:
Now let's find the slope of the chord of contact of point P with respect to the hyperbola
Slope =$- \frac{{\left( {{\text{coefficient of }}x} \right)}}{{\left( {{\text{coefficient of }}y} \right)}}$
Slope of equation (1) is given by
$\Rightarrow {m_1} = \frac{{ - \frac{h}{{{a^2}}}}}{{ - \frac{y}{{{b^2}}}}} \\\ \Rightarrow {m_1} = \frac{h}{{{a^2}}}*\frac{{{b^2}}}{y} \\\$
Step 3:
Now the chord of contact of a point $\left( {{x_1},{y_1}} \right)$ with respect to its auxiliary circle is given by
$x{x_1} + y{y_1} = {a^2} - {b^2}$
So the chord of contact of a point (h,k) with respect to its auxiliary circle is given by
$xh + yk = {a^2} - {b^2}$………..(2)
Step 4:
Now let's find the slope of the chord of contact of point P with respect to its auxiliary circle
Slope =$- \frac{{\left( {{\text{coefficient of }}x} \right)}}{{\left( {{\text{coefficient of }}y} \right)}}$
Slope of equation (2) is given by
$\Rightarrow {m_2} = \frac{{ - h}}{k}$
Step 5:
We are given chords of contact of a point P w.r.t. a hyperbola and its auxiliary circle are at right angles.
It means that the product of their slopes is – 1

$$\Rightarrow {m_1}{m_2} = - 1 \\\ \Rightarrow \frac{h}{{{a^2}}}*\frac{{{b^2}}}{k}*\frac{{ - h}}{k} = - 1 \\\ \Rightarrow \frac{{ - {h^2}{b^2}}}{{{k^2}{a^2}}} = - 1 \\\ \Rightarrow \frac{{{h^2}{b^2}}}{{{k^2}{a^2}}} = 1 \\\ \Rightarrow {h^2}{b^2} = {k^2}{a^2} \\\ \Rightarrow {h^2}{b^2} - {k^2}{a^2} = 0 \\\ ..$$

Step 6 :
Let's divide the above equation by ${a^2}{b^2}$

$$\Rightarrow \frac{{{h^2}{b^2}}}{{{a^2}{b^2}}} - \frac{{{k^2}{a^2}}}{{{a^2}{b^2}}} = 0 \\\ \Rightarrow \frac{{{h^2}}}{{{a^2}}} - \frac{{{k^2}}}{{{b^2}}} = 0 \\\$$

This is equation of the asymptotes of the hyperbola$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$
Therefore from this we get that the point lies on one of the asymptotes.

The correct option is C.

Note: Equilateral hyperbola = rectangular hyperbola.
If a hyperbola is equilateral then the conjugate hyperbola is also equilateral.
A hyperbola and its conjugate have the same asymptote.
The equation of the pair of asymptotes differ from the hyperbola & the conjugate hyperbola by the same constant only.