Question

Find the locus of the poles of normal chords of the hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$
A. ${a^2}{x^2} - {b^2}{y^2} = {a^4} + {b^4}$
B. ${a^6}{y^2} - {b^6}{x^2} = {\left( {{a^2} + {b^2}} \right)^2}{x^2}{y^2}$
C. ${a^2}{y^2} - {b^2}{x^2} = {\left( {{a^2} + {b^2}} \right)^2}xy$
D. None of these

Answer 1

Hint:
Here, we will use the equation of the hyperbola and the equation of the normal chord of a hyperbola to solve this question. We will find the equation of the polar form with respect to the normal equation of the given hyperbola. Then comparing the coefficients we will be able to solve it further and hence, find the locus of the poles of normal chords of the given hyperbola.

Formula Used:
We will use the following formulas:
1. Equation of normal of hyperbola, $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$
2. The equation of a normal chord of given hyperbola, $ax\cos \theta + by\cot \theta = {a^2} + {b^2}$
3. $\cos \theta = \dfrac{1}{{\sec \theta }}$
4. $\cot \theta = \dfrac{1}{{\tan \theta }}$

Complete step by step solution:
Let $\left( {h,k} \right)$ be the pole of a normal chord of the given hyperbola.
Now, we know that equation of any normal to the hyperbola is:
$\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$……………………………..$\left( 1 \right)$
Also, the equation of a normal chord of the given hyperbola is:
$ax\cos \theta + by\cot \theta = {a^2} + {b^2}$
As we know, $\cos \theta = \dfrac{1}{{\sec \theta }}$ and $\cot \theta = \dfrac{1}{{\tan \theta }}$.
Hence, this equation can also be written as:
$\dfrac{{ax}}{{\sec \theta }} + \dfrac{{by}}{{\tan \theta }} = {a^2} + {b^2}$…………………………….$\left( 2 \right)$
Now, the equation of the polar of $\left( {h,k} \right)$ with respect to the normal equation of the given hyperbola, i.e. from $\left( 1 \right)$, we get,
$\dfrac{{xh}}{{{a^2}}} - \dfrac{{yk}}{{{b^2}}} = 1$……………………………$\left( 3 \right)$
Now, as $\left( 2 \right)$ and $\left( 3 \right)$ are the equations of the same straight line and both represent the polar of $\left( {h,k} \right)$
Hence, comparing their coefficients, we get,
$\dfrac{{\dfrac{h}{{{a^2}}}}}{{\dfrac{a}{{\sec \theta }}}} = \dfrac{{\dfrac{k}{{{b^2}}}}}{{\dfrac{b}{{\tan \theta }}}} = \dfrac{1}{{{a^2} + {b^2}}}$
\[ \Rightarrow \dfrac{{h\sec \theta }}{{{a^3}}} = \dfrac{{k\tan \theta }}{{{b^3}}} = \dfrac{1}{{{a^2} + {b^2}}}\]
Hence, we get,
\[\sec \theta = \dfrac{{{a^3}}}{{h\left( {{a^2} + {b^2}} \right)}}\]
\[\tan \theta = \dfrac{{{b^3}}}{{k\left( {{a^2} + {b^2}} \right)}}\]
Now, we know that, ${\sec ^2}\theta - {\tan ^2}\theta = 1$.
Hence, substituting the above values, we get,
\[ \Rightarrow {\left( {\dfrac{{{a^3}}}{{h\left( {{a^2} + {b^2}} \right)}}} \right)^2} - {\left( {\dfrac{{{b^3}}}{{k\left( {{a^2} + {b^2}} \right)}}} \right)^2} = 1\]
Applying the exponent on the terms, we get
\[ \Rightarrow \dfrac{{{a^6}}}{{{h^2}{{\left( {{a^2} + {b^2}} \right)}^2}}} - \dfrac{{{b^6}}}{{{k^2}{{\left( {{a^2} + {b^2}} \right)}^2}}} = 1\]
Solving further, we get,
\[ \Rightarrow \dfrac{{{a^6}}}{{{h^2}}} - \dfrac{{{b^6}}}{{{k^2}}} = {\left( {{a^2} + {b^2}} \right)^2}\]
Again substituting $\left( {h,k} \right) = \left( {x,y} \right)$, we get,
\[ \Rightarrow \dfrac{{{a^6}}}{{{x^2}}} - \dfrac{{{b^6}}}{{{y^2}}} = {\left( {{a^2} + {b^2}} \right)^2}\]
Hence, locus of $\left( {h,k} \right)$ is:
\[ \Rightarrow \dfrac{{{a^6}}}{{{x^2}}} - \dfrac{{{b^6}}}{{{y^2}}} = {\left( {{a^2} + {b^2}} \right)^2}\]
Taking LCM on the LHS,
\[ \Rightarrow \dfrac{{{a^6}{y^2} - {b^6}{x^2}}}{{{x^2}{y^2}}} = {\left( {{a^2} + {b^2}} \right)^2}\]
Taking denominator to the RHS, we get,
\[ \Rightarrow {a^6}{y^2} - {b^6}{x^2} = {\left( {{a^2} + {b^2}} \right)^2}{x^2}{y^2}\]

Therefore, the locus of the poles of normal chords of the hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ is ${a^6}{y^2} - {b^6}{x^2} = {\left( {{a^2} + {b^2}} \right)^2}{x^2}{y^2}$
Hence, option B is the correct answer.

Note:
In mathematics, hyperbola is a symmetrical open curve which is formed by the intersection of a circular cone with a plane at a smaller angle with its axis than the side of the coin. It is sometimes defined as the difference of distances between a set of points, which are present in a plane to two fixed points and is a given positive constant. A locus is a set of all points that satisfy some properties which usually results as a curve or a surface. In order to solve this question, it is really important to have this knowledge about hyperbola and the locus respectively.