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# Find the points common to the hyperbola $25{{\text{x}}^2} - {\text{9}}{{\text{y}}^2} = 225$ and the straight line $25{\text{x + 12y - 45 = 0}}$.

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Hint:– Transform the equation of straight line such that the variable x is in terms of variable y. Substitute y in the equation of hyperbola. Obtain the value of y and substitute for x.

Complete step-by-step solution -
Given Data –
Equation of hyperbola $25{{\text{x}}^2} - {\text{9}}{{\text{y}}^2} = 225$ and Equation of straight line $25{\text{x + 12y - 45 = 0}}$.
Rewriting the equation of straight line, we get ${\text{x = }}\dfrac{{45 - 12{\text{y}}}}{{25}}{\text{ }} \to {\text{Equation 1}}$.
We found the value of x in terms of y.
Now substitute this x in the equation of hyperbola, we get
${\text{25}}{\left( {\dfrac{{45 - 12{\text{y}}}}{{25}}} \right)^2} - 9{{\text{y}}^2} = 225$
$\Rightarrow 2025{\text{ + 144}}{{\text{y}}^2} - 1080{\text{y - 225}}{{\text{y}}^2} = 5625 \\ \Rightarrow 81{{\text{y}}^2} + 1080{\text{y + 3600 = 0}} \\$
Upon simplifying,
$\Rightarrow {\text{9}}{{\text{y}}^2} + 120{\text{y + 400 = 0}}$
Now solve this equation to get the values of y.
$\Rightarrow {\left( {{\text{3y + 20}}} \right)^2} = 0 \\ \Rightarrow \left( {{\text{3y + 20}}} \right) = 0 \\ \Rightarrow {\text{y = }}\dfrac{{ - 20}}{3} \\$
We have obtained the value of y, substituting this value in the Equation 1 gives us the value of x.
${\text{x = }}\dfrac{{45 - 12\left( {\dfrac{{ - 20}}{3}} \right)}}{{25}} = \dfrac{{45{\text{ + 80}}}}{{25}} = 5 \\ \left( {{\text{x,y}}} \right) = \left( {{\text{5,}}\dfrac{{ - 20}}{3}} \right) \\$
The common point to the hyperbola and the straight line is$\left( {{\text{5,}}\dfrac{{ - 20}}{3}} \right)$.
Note:– In order to solve this type of question the key is to transform one of the equations such that we have one variable in terms of another. Then the other equation reduces into a single variable equation. On finding the value of one variable the other can be found simply by substituting.
Last updated date: 19th Sep 2023
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