
Find the hyperbola whose asymptotes are given $2x - y = 3$, $3x + y = 7$ and which passes through the point $\left( {1,1} \right)$.
Answer
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Hint: In the given question, we have to find out the equation of hyperbola and information of asymptote its equation is given as well as hyperbola passes through is $\left( {1,1} \right)$ also given. To work it out, use the relation (product of asymptote $ + $constant $ = 0$) as it is also an equation of hyperbola. Since, $\left( {1,1} \right)$ is a points from where it passes put $\left( {x,y} \right) = \left( {1,1} \right)$ in the said equation & find value of $k$. Then put the value of $k$ in the equation & work out the equation of hyperbola.
Complete step-by-step answer:
We know that –
Equation of hyperbola differs from the joint equation of its asymptotes by a constant.
Equation of hyperbola –
$ \Rightarrow $(product of asymptote) $ + $constant$\left( k \right)$$ \Rightarrow $$0$
Given –
Asymptote equations are –
${
2x - y = 3 \\
\Rightarrow 2x - y - 3 = 0 \\
\therefore \left( {2x - y - 3} \right)\left( {3x + y - 7} \right) + k = 0..........(i) \\
} $
And,
${
3x + y = 7 \\
\Rightarrow 3x + y - 7 = 0 \\
} $
Now, it is given that hyperbola passes through $\left( {1,1} \right)$
$\therefore $Putting $\left( {x.y} \right) = \left( {1,1} \right)$ in equation $(i)$
${
\Rightarrow (2 - 1 - 3)(3 + 1 - 7) + k = 0 \\
\Rightarrow ( - 2 \times - 3) + k = 0 \\
\Rightarrow k = - 6 \\
} $
$\therefore $Putting $k = - 6$ in require hyperbola equation, i.e.
${
\Rightarrow (2x - y - 3)(3x + y - 7) + k = 0 \\
\Rightarrow (2x - y - 3)\left( {3x + y - 7} \right) - 6 = 0 \\
\Rightarrow 6{x^2} + 2xy - 14x - 3xy - {y^2} + 7y - 9x - 3y + 21 - 6 = 0 \\
} $
$ \Rightarrow 6{x^2} - {y^2} - xy - 23x + 4y + 15 = 0$
Hence $6{x^2} - {y^2} - xy - 23x + 4y + 15 = 0$ is the required hyperbola equation.
Note: The question is related to the coordinate chapter and conic section. The Conic section is a huge chapter including the topics of circle, ellipse, parabola, hyperbola. To solve questions related to them, one should know each and every point of their properties and terms.
Hyperbola is a collection of points in a plane such that there is a constant distance between two fixed points and each point. Hyperbola is made up of two similar curves that resemble a parabola.
Complete step-by-step answer:
We know that –
Equation of hyperbola differs from the joint equation of its asymptotes by a constant.
Equation of hyperbola –
$ \Rightarrow $(product of asymptote) $ + $constant$\left( k \right)$$ \Rightarrow $$0$
Given –
Asymptote equations are –
${
2x - y = 3 \\
\Rightarrow 2x - y - 3 = 0 \\
\therefore \left( {2x - y - 3} \right)\left( {3x + y - 7} \right) + k = 0..........(i) \\
} $
And,
${
3x + y = 7 \\
\Rightarrow 3x + y - 7 = 0 \\
} $
Now, it is given that hyperbola passes through $\left( {1,1} \right)$
$\therefore $Putting $\left( {x.y} \right) = \left( {1,1} \right)$ in equation $(i)$
${
\Rightarrow (2 - 1 - 3)(3 + 1 - 7) + k = 0 \\
\Rightarrow ( - 2 \times - 3) + k = 0 \\
\Rightarrow k = - 6 \\
} $
$\therefore $Putting $k = - 6$ in require hyperbola equation, i.e.
${
\Rightarrow (2x - y - 3)(3x + y - 7) + k = 0 \\
\Rightarrow (2x - y - 3)\left( {3x + y - 7} \right) - 6 = 0 \\
\Rightarrow 6{x^2} + 2xy - 14x - 3xy - {y^2} + 7y - 9x - 3y + 21 - 6 = 0 \\
} $
$ \Rightarrow 6{x^2} - {y^2} - xy - 23x + 4y + 15 = 0$
Hence $6{x^2} - {y^2} - xy - 23x + 4y + 15 = 0$ is the required hyperbola equation.
Note: The question is related to the coordinate chapter and conic section. The Conic section is a huge chapter including the topics of circle, ellipse, parabola, hyperbola. To solve questions related to them, one should know each and every point of their properties and terms.
Hyperbola is a collection of points in a plane such that there is a constant distance between two fixed points and each point. Hyperbola is made up of two similar curves that resemble a parabola.
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