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Question: Find the hyperbola whose asymptotes are given \(2x - y = 3\), \(3x + y = 7\) and which passes throug...

Find the hyperbola whose asymptotes are given 2xy=32x - y = 3, 3x+y=73x + y = 7 and which passes through the point (1,1)\left( {1,1} \right).

Explanation

Solution

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 (1,1)\left( {1,1} \right) also given. To work it out, use the relation (product of asymptote ++constant =0 = 0) as it is also an equation of hyperbola. Since, (1,1)\left( {1,1} \right) is a points from where it passes put (x,y)=(1,1)\left( {x,y} \right) = \left( {1,1} \right) in the said equation & find value of kk. Then put the value of kk 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(k)\left( k \right) \Rightarrow 00
Given –
Asymptote equations are –
2xy=3 2xy3=0 (2xy3)(3x+y7)+k=0..........(i) { 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 3x+y7=0 { 3x + y = 7 \\\ \Rightarrow 3x + y - 7 = 0 \\\ }
Now, it is given that hyperbola passes through (1,1)\left( {1,1} \right)
\therefore Putting (x.y)=(1,1)\left( {x.y} \right) = \left( {1,1} \right) in equation (i)(i)
(213)(3+17)+k=0 (2×3)+k=0 k=6 { \Rightarrow (2 - 1 - 3)(3 + 1 - 7) + k = 0 \\\ \Rightarrow ( - 2 \times - 3) + k = 0 \\\ \Rightarrow k = - 6 \\\ }
\therefore Putting k=6k = - 6 in require hyperbola equation, i.e.
(2xy3)(3x+y7)+k=0 (2xy3)(3x+y7)6=0 6x2+2xy14x3xyy2+7y9x3y+216=0 { \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 \\\ }
6x2y2xy23x+4y+15=0\Rightarrow 6{x^2} - {y^2} - xy - 23x + 4y + 15 = 0

Hence 6x2y2xy23x+4y+15=06{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.