Question

Tangents are drawn from any point on the hyperbola $\dfrac{{{x}^{2}}}{9}-\dfrac{{{y}^{2}}}{4}=1$ to the circle ${{x}^{2}}+{{y}^{2}}=9$. Find the locus of midpoint of the chord of contact.

Answer 1

First, assume any point on the hyperbola to the circle. Then, the chord of contact of the circle concerning the point. After that find the equation of chord in mid-point form. Now equate both the equations to get the value of $\sec \theta$ and $\tan \theta$. As we know ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$, substitute the values in it and simplify. The equation derived is the locus of midpoint of the chord of contact.

Complete step-by-step solution:
Given: - Equation of the hyperbola is $\dfrac{{{x}^{2}}}{9}-\dfrac{{{y}^{2}}}{4}=1$.
The equation of the circle is ${{x}^{2}}+{{y}^{2}}=9$.
Let any point on the hyperbola to the circle be $\left( 3\sec \theta ,2\tan \theta \right)$ and the midpoint of the chord of contact be $\left( {{x}_{1}},{{y}_{1}} \right)$.
Then, the chord of contact of the circle concerning the point $\left( 3\sec \theta ,2\tan \theta \right)$ is,
$\left( 3\sec \theta \right)x+\left( 2\tan \theta \right)y=9$...............….. (1)
Now, the equation of chord in mid-point form is
$x{{x}_{1}}+y{{y}_{1}}={{x}_{1}}^{2}+{{y}_{1}}^{2}$.................….. (2)
Since both the equation (1) and (2) represent the same line.
As we know that for unique or many solutions,
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$
Then from equation (1) and (2),
$\Rightarrow$ $\dfrac{3\sec\theta}{{{x}_{1}}}=\dfrac{2\tan\theta}{{{y}_{1}}}=\dfrac{9}{{{x}_{1}}^{2}+{{y}_{1}}^{2}}$
Now, take the first and last term to find the value of $\sec \theta$,
$\Rightarrow$ $\dfrac{3\sec \theta }{{{x}_{1}}}=\dfrac{9}{{{x}_{1}}^{2}+{{y}_{1}}^{2}}$
Multiply the denominator of the left side to the numerator of the right side,
$\Rightarrow$ $3\sec \theta =\dfrac{9{{x}_{1}}}{{{x}_{1}}^{2}+{{y}_{1}}^{2}}$
Divide both sides by 3,
$\Rightarrow$ $\sec \theta =\dfrac{3{{x}_{1}}}{{{x}_{1}}^{2}+{{y}_{1}}^{2}}$
Now, take the second and last term to find the value of $\tan \theta$,
$\Rightarrow$ $\dfrac{2\tan \theta }{{{y}_{1}}}=\dfrac{9}{{{x}_{1}}^{2}+{{y}_{1}}^{2}}$
Multiply the denominator of the left side to the numerator of the right side,
$\Rightarrow$ $2\tan \theta =\dfrac{9{{y}_{1}}}{{{x}_{1}}^{2}+{{y}_{1}}^{2}}$
Divide both sides by 2,
$\Rightarrow$ $\tan \theta =\dfrac{9{{y}_{1}}}{2\left( {{x}_{1}}^{2}+{{y}_{1}}^{2} \right)}$
As we know, ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$.
Substitute the values of $\sec \theta$ and $\tan \theta$ from above,
$\Rightarrow$ ${{\left( \dfrac{3{{x}_{1}}}{{{x}_{1}}^{2}+{{y}_{1}}^{2}} \right)}^{2}}-{{\left[ \dfrac{9{{y}_{1}}}{2\left( {{x}_{1}}^{2}+{{y}_{1}}^{2} \right)} \right]}^{2}}=1$
Square the terms,
$\Rightarrow$ $\dfrac{9{{x}_{1}}^{2}}{{{\left({{x}_{1}}^{2}+{{y}_{1}}^{2}\right)}^{2}}}-\dfrac{81{{y}_{1}}^{2}}{4{{\left( {{x}_{1}}^{2}+{{y}_{1}}^{2} \right)}^{2}}}=1$
Take ${{\left( {{x}_{1}}^{2}+{{y}_{1}}^{2} \right)}^{2}}$ common from the denominator,
$\Rightarrow$ $\dfrac{1}{{{\left( {{x}_{1}}^{2}+{{y}_{1}}^{2} \right)}^{2}}}\left[ 9{{x}_{1}}^{2}-\dfrac{81{{y}_{1}}^{2}}{4} \right]=1$
Divide both sides by $\dfrac{{{\left( {{x}_{1}}^{2}+{{y}_{1}}^{2} \right)}^{2}}}{81}$,
$\Rightarrow$ $\dfrac{9{{x}_{1}}^{2}}{81}-\dfrac{81{{y}_{1}}^{2}}{4\times 81}=\dfrac{{{\left( {{x}_{1}}^{2}+{{y}_{1}}^{2} \right)}^{2}}}{81}$
Cancel out the common factors from numerator and denominator,
$\Rightarrow$ $\dfrac{{{x}_{1}}^{2}}{9}-\dfrac{{{y}_{1}}^{2}}{4}={{\left( \dfrac{{{x}_{1}}^{2}+{{y}_{1}}^{2}}{9} \right)}^{2}}$

Hence, the locus of midpoint of the chord of contact is $\dfrac{{{x}_{1}}^{2}}{9}-\dfrac{{{y}_{1}}^{2}}{4}={{\left( \dfrac{{{x}_{1}}^{2}+{{y}_{1}}^{2}}{9} \right)}^{2}}$.

Note: A hyperbola is an open curve with two branches, the intersection of a plane with both halves of a double cone. The plane does not have to be parallel to the axis of the cone; the hyperbola will be symmetrical in any case.
The standard form of the equation of the hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.