Question

Tangents are drawn to the hyperbola $\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{4} = 1$, parallel to the straight line $2x - y = 1$. the points of contact of the tangents on the hyperbola are:
A. $\left( {\dfrac{9}{{\sqrt 2 }},\dfrac{1}{{\sqrt 2 }}} \right)$
B. $\left( { - \dfrac{9}{{2\sqrt 2 }}, - \dfrac{1}{{\sqrt 2 }}} \right)$
C. $\left( {3\sqrt 3 , - 2\sqrt 2 } \right)$
D. $\left( { - 3\sqrt 3 ,2\sqrt 2 } \right)$

Answer 1

Hint: Take the derivative of both the equations i.e the equation of straight line and equation of hyperbola and by rearranging the equations, determine the value of $x$. After finding the value of $x$ substitute the value of $x$ in the straight line to get the value of $y$.

Complete step-by-step answer:
Given data:
The hyperbola equation is $\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{4} = 1$
The straight-line equation is $2x - y = 1$
Now, calculate the slope of the straight line by differentiating the equation $2x - y = 1$ with respect to $x$.
\[
\dfrac{d}{{dx}}\left( {2x - y} \right) = \dfrac{d}{{dx}}\left( 1 \right)\\
2 - \dfrac{{dy}}{{dx}} = 0\\
\dfrac{{dy}}{{dx}} = 2
\]
Now, differentiate the hyperbola equation $\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{4} = 1$ with respect to $x$ :
$
\dfrac{d}{{dx}}\left( {\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{4}} \right) = \dfrac{d}{{dx}}\left( 1 \right)\\
\dfrac{{2x}}{9} - \dfrac{{2y}}{4}\dfrac{{dy}}{{dx}} = 0\\
\dfrac{{dy}}{{dx}} = \dfrac{{2x}}{9} \times \dfrac{4}{{2y}}\\
 = \dfrac{{4x}}{{9y}}
$
Substitute the value of $\dfrac{{dy}}{{dx}} = 2$ in $\dfrac{{dy}}{{dx}} = \dfrac{{4x}}{{9y}}$.
$
2 = \dfrac{{4x}}{{9y}}\\
4x = 18y\\
x = \dfrac{{18}}{4}y\\
x = \dfrac{9}{2}y
$
Substitute the value of$x$ in $\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{4} = 1$.
$
\dfrac{{{{\left( {\dfrac{9}{2}y} \right)}^2}}}{9} - \dfrac{{{y^2}}}{4} = 1\\
\dfrac{{9{y^2}}}{4} - \dfrac{{{y^2}}}{4} = 1\\
\dfrac{{8{y^2}}}{4} = 1\\
y = \pm \dfrac{1}{{\sqrt 2 }}
$
Now, calculate the value of $x$ by substituting the value of y in $x = \dfrac{9}{2}y$.
$
x = \dfrac{9}{2}\left( { \pm \dfrac{1}{{\sqrt 2 }}} \right)\\
 = \pm \dfrac{9}{{2\sqrt 2 }}
$
Hence, the points of contact are $\left( {\dfrac{9}{{2\sqrt 2 }},\dfrac{1}{{\sqrt 2 }}} \right){\rm{ and }}\left( { - \dfrac{9}{{2\sqrt 2 }}, - \dfrac{1}{{\sqrt 2 }}} \right)$.

Option (A) and (B) are the correct answers.

Note: The general equation of the tangent to hyperbola is $y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} $, where the slope is given by $m$. Make sure that chain rule is used in derivatives of complex functions.