Question

If the line $lx+my+n=0$ is a normal to the hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, then show that $\dfrac{{{a}^{2}}}{{{l}^{2}}}-\dfrac{{{b}^{2}}}{{{m}^{2}}}={{\left( \dfrac{{{a}^{2}}+{{b}^{2}}}{n} \right)}^{2}}$?

Answer 1

We start solving the problem by using the fact that the parametric equation of the normal of the hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $\dfrac{ax}{\sec \theta }+\dfrac{by}{\tan \theta }={{a}^{2}}+{{b}^{2}}$ and rearrange the terms in it to get the equation in the form of $px+qy+r=0$. We then compare the coefficients of x, y and constant terms of the obtained equation of normal with $lx+my+n=0$ to find the values of l, m and n. We then assume $\dfrac{{{a}^{2}}}{{{l}^{2}}}-\dfrac{{{b}^{2}}}{{{m}^{2}}}$ and substitute the values of l and m and make necessary calculations to get the required result.

Complete step by step answer:
According to the problem, we are given that the line $lx+my+n=0$ is a normal to the hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$. We need to show that $\dfrac{{{a}^{2}}}{{{l}^{2}}}-\dfrac{{{b}^{2}}}{{{m}^{2}}}={{\left( \dfrac{{{a}^{2}}+{{b}^{2}}}{n} \right)}^{2}}$.

We know that the parametric equation of the normal of the hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $\dfrac{ax}{\sec \theta }+\dfrac{by}{\tan \theta }={{a}^{2}}+{{b}^{2}}$. Let us rewrite this equation in the form of $px+qy+r=0$.
So, we get the equation of the normal as $\dfrac{ax}{\sec \theta }+\dfrac{by}{\tan \theta }-\left( {{a}^{2}}+{{b}^{2}} \right)=0$---(1).
According to the problem, we are given that $lx+my+n=0$ is normal to the hyperbola. So, let us compare this equation with equation (1).
So, we get $l=\dfrac{a}{\sec \theta }$, $m=\dfrac{b}{\tan \theta }$ and $n=-\left( {{a}^{2}}+{{b}^{2}} \right)$ ---(2).
Let us consider $\dfrac{{{a}^{2}}}{{{l}^{2}}}-\dfrac{{{b}^{2}}}{{{m}^{2}}}$. Let us substitute the results obtained from equation (2) in this.
$\Rightarrow \dfrac{{{a}^{2}}}{{{l}^{2}}}-\dfrac{{{b}^{2}}}{{{m}^{2}}}=\dfrac{{{a}^{2}}}{{{\left( \dfrac{a}{\sec \theta } \right)}^{2}}}-\dfrac{{{b}^{2}}}{{{\left( \dfrac{b}{\tan \theta } \right)}^{2}}}$.
$\Rightarrow \dfrac{{{a}^{2}}}{{{l}^{2}}}-\dfrac{{{b}^{2}}}{{{m}^{2}}}=\dfrac{{{a}^{2}}}{\dfrac{{{a}^{2}}}{{{\sec }^{2}}\theta }}-\dfrac{{{b}^{2}}}{\dfrac{{{b}^{2}}}{{{\tan }^{2}}\theta }}$.
$\Rightarrow \dfrac{{{a}^{2}}}{{{l}^{2}}}-\dfrac{{{b}^{2}}}{{{m}^{2}}}={{\sec }^{2}}\theta -{{\tan }^{2}}\theta$.
We know that ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$.
$\Rightarrow \dfrac{{{a}^{2}}}{{{l}^{2}}}-\dfrac{{{b}^{2}}}{{{m}^{2}}}=1$ ---(3).
We know that ${{\left( -1 \right)}^{2}}=1$. Let us use this in equation (2).
$\Rightarrow \dfrac{{{a}^{2}}}{{{l}^{2}}}-\dfrac{{{b}^{2}}}{{{m}^{2}}}={{\left( -1 \right)}^{2}}$.
$\Rightarrow \dfrac{{{a}^{2}}}{{{l}^{2}}}-\dfrac{{{b}^{2}}}{{{m}^{2}}}={{\left( \dfrac{{{a}^{2}}+{{b}^{2}}}{-\left( {{a}^{2}}+{{b}^{2}} \right)} \right)}^{2}}$.
From equation (2), we get
$\Rightarrow \dfrac{{{a}^{2}}}{{{l}^{2}}}-\dfrac{{{b}^{2}}}{{{m}^{2}}}={{\left( \dfrac{{{a}^{2}}+{{b}^{2}}}{n} \right)}^{2}}$.
So, we have proved $\dfrac{{{a}^{2}}}{{{l}^{2}}}-\dfrac{{{b}^{2}}}{{{m}^{2}}}={{\left( \dfrac{{{a}^{2}}+{{b}^{2}}}{n} \right)}^{2}}$.

Note: We can also find the slope of the normal first by using the facts that the point on hyperbola is of form $\left( a\sec \theta ,b\tan \theta \right)$ and slope of the tangent by using ${{\left. \dfrac{dy}{dx} \right|}_{\left( a\sec \theta ,b\tan \theta \right)}}$. We should not make calculation mistakes while solving this problem. We can also use the fact that if two lines ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ are identical to each other, then $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$ to solve the problem.