Question

Find the hyperbola whose asymptotes are given $2x - y = 3$ , $3x + y = 7$ and which passes through point $\left( {1,1} \right)$

Answer 1

Hint : In order to determine the equation of the hyperbola having asymptotes $2x - y = 3$ , $3x + y = 7$ and passes through the point $\left( {1,1} \right)$ , use the fact that the equation of any hyperbola differs from the joint equation of its asymptotes by a constant. Let the constant be k and form the equation as $\left( {2x - y - 3} \right)\left( {3x + y - 7} \right) + k = 0$ . Determine the value of k by putting the point $\left( {1,1} \right)$ ,and put the value of k back into the equation. Expand and combine all the like terms to get your required equation of hyperbola.

Complete step-by-step answer :
We are given hyperbola whose asymptotes are $2x - y = 3$ , $3x + y = 7$ which passes through the point $\left( {1,1} \right)$ , and we have to find the equation of this hyperbola
First we know that the equation of hyperbola differs from the joint equation of its asymptotes by a constant.
Let the constant be $k$ ,
In mathematical expression we can say that
Equation of Hyperbola $= \left( {2x - y - 3} \right)\left( {3x + y - 7} \right) + k = 0$ --------(1)
According to the question, the hyperbola passes through the point $\left( {1,1} \right)$ which means the point $\left( {1,1} \right)$ satisfies the equation of hyperbola.
Finding the value of constant $k$ from the equation by putting $x = 1$ and $y = 1$ , equation becomes
$\Rightarrow \left( {2\left( 1 \right) - \left( 1 \right) - 3} \right)\left( {3\left( 1 \right) + \left( 1 \right) - 7} \right) + k = 0 \\\ \Rightarrow \left( {2 - 1 - 3} \right)\left( {3 + 1 - 7} \right) + k = 0 \\\ \Rightarrow \left( { - 2} \right)\left( { - 3} \right) + k = 0 \\\ \Rightarrow 6 + k = 0 \\\ \Rightarrow k = - 6 \;$
So the required equation of hyperbola now becomes by putting $k = - 6$ in equation(1)
Equation of Hyperbola $= \left( {2x - y - 3} \right)\left( {3x + y - 7} \right) + k = 0$
$= \left( {2x - y - 3} \right)\left( {3x + y - 7} \right) - 6 = 0$
Now expanding the bracket, we get
$= \left( {2x} \right)\left( {3x + y - 7} \right) - y\left( {3x + y - 7} \right) - 3\left( {3x + y - 7} \right) - 6 = 0 \\\ = 6{x^2} + 2xy - 14x - 3xy - {y^2} + 7y - 9x - 3y + 21 - 6 = 0 \;$
Combine all the like terms, we get
$= 6{x^2} - xy - {y^2} - 23x + 4y + 15$
Therefore, Equation of Hyperbola $= 6{x^2} - xy - {y^2} - 23x + 4y + 15$ .
So, the correct answer is “ $6{x^2} - xy - {y^2} - 23x + 4y + 15$ ”.

Note : 1. When the centre of hyper is at the origin and foci are on the x-axis or y-axis , the Standard equation of hyperbola is
$\left[ {\left( {\dfrac{{{x^2}}}{{{a^2}}}} \right) - \left( {\dfrac{{{y^2}}}{{{b^2}}}} \right)} \right] = 1$
2.Make sure that the expansion of the terms is done carefully while determining the equation.