
The centre of the conic represented by the equation$2{x^2} - 72xy + 23{y^2} - 4x - 28y - 48 = 0$ is
A. $(\dfrac{{11}}{{15}},\dfrac{2}{{25}})$
B. $(\dfrac{{11}}{{25}}, - \dfrac{2}{{25}})$
C. $( - \dfrac{{11}}{{25}},\dfrac{2}{{25}})$
D. $( - \dfrac{{11}}{{25}}, - \dfrac{2}{{25}})$
Answer
164.7k+ views
Hint:
In this problem, we have been provided that the conic is represented by the equation\[2{x^2} - 72xy + 23{y^2} - 4x - 28y - 48 = 0\] and we are asked to determine the centre of the conic. To determine the centre of the conic section, we have to use the general formula of the conic section and we will obtain the value of coefficients and by substituting the values on given conditions, we will obtain a desired solution.
Formula used:
Compare the given equation to the conic section's general form. \[a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\]
Find the coefficients' values. Then, to determine the centre, resolve the following equations:\[a{x_1} + h{y_1} + g = 0\]and \[h{x_1} + b{y_1} + f = 0\]
Complete step-by-step solution:
A conic section, or simply a conic, is the locus of a moving point if it moves in a plane such that its distance from a fixed point always bears a fixed straight-line distance divided by a constant.
A hyperbola is a collection of all locations (x, y) where the difference between their respective distances from other points is always constant. A hyperbola is the name for the fixed point of the foci. There are two branches in the hyperbola graph. Each component has a parabola-like appearance but a marginally different shape. Two of a hyperbola's vertices are located on the transverse axis, which is its axis of symmetry. The hyperbola's transverse axis can be either horizontal or vertical.
We are clear that a conic section has the general form:
\[a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\]-------- (1)
Let the conic's centre be at \[({x_1},{y_1})\]. Consequently, the following equations will be satisfied by the conic's centre:
\[a{x_1} + h{y_1} + g = 0\]and \[h{x_1} + b{y_1} + f = 0\]
Now the given equation is: \[2{x^2} - 72xy + 23{y^2} - 4x - 28y - 48 = 0\]------- (2)
Now we can obtain by comparing the coefficients of the variables from (1) and (2),
\[a = 2\]
\[b = 23\]
\[c = - 48\]
\[2h = - 72 \Rightarrow h = - 36\]
\[2g = - 4 \Rightarrow g = - 2\]
\[2f = - 28 \Rightarrow f = - 14\]
Let us put the values of the coefficients in our condition.
\[2{x_1} - 36{y_1} - 2 = 0\]
\[ - 36{x_1} + 23{y_1} - 14 = 0\]
Solving above equation we get
\[({x_1},{y_1}) = \left( { - \frac{{11}}{{25}}, - \frac{2}{{25}}} \right)\]
Therefore, the centre of the conic represented by the equation is
\[\left( {\frac{{11}}{{25}}, - \frac{2}{{25}}} \right)\]
Hence option D is correct
Note: We can also find out centre of a conic by the following formula: \[{x_1} = \frac{{hf - bg}}{{ab - {h^2}}},{y_1} = \frac{{gh - af}}{{ab - {h^2}}}\].
Students often make mistakes in these types of problems because these types of problems include many formulas to remember and we should apply the formula to get the desired answer.
In this problem, we have been provided that the conic is represented by the equation\[2{x^2} - 72xy + 23{y^2} - 4x - 28y - 48 = 0\] and we are asked to determine the centre of the conic. To determine the centre of the conic section, we have to use the general formula of the conic section and we will obtain the value of coefficients and by substituting the values on given conditions, we will obtain a desired solution.
Formula used:
Compare the given equation to the conic section's general form. \[a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\]
Find the coefficients' values. Then, to determine the centre, resolve the following equations:\[a{x_1} + h{y_1} + g = 0\]and \[h{x_1} + b{y_1} + f = 0\]
Complete step-by-step solution:
A conic section, or simply a conic, is the locus of a moving point if it moves in a plane such that its distance from a fixed point always bears a fixed straight-line distance divided by a constant.
A hyperbola is a collection of all locations (x, y) where the difference between their respective distances from other points is always constant. A hyperbola is the name for the fixed point of the foci. There are two branches in the hyperbola graph. Each component has a parabola-like appearance but a marginally different shape. Two of a hyperbola's vertices are located on the transverse axis, which is its axis of symmetry. The hyperbola's transverse axis can be either horizontal or vertical.
We are clear that a conic section has the general form:
\[a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0\]-------- (1)
Let the conic's centre be at \[({x_1},{y_1})\]. Consequently, the following equations will be satisfied by the conic's centre:
\[a{x_1} + h{y_1} + g = 0\]and \[h{x_1} + b{y_1} + f = 0\]
Now the given equation is: \[2{x^2} - 72xy + 23{y^2} - 4x - 28y - 48 = 0\]------- (2)
Now we can obtain by comparing the coefficients of the variables from (1) and (2),
\[a = 2\]
\[b = 23\]
\[c = - 48\]
\[2h = - 72 \Rightarrow h = - 36\]
\[2g = - 4 \Rightarrow g = - 2\]
\[2f = - 28 \Rightarrow f = - 14\]
Let us put the values of the coefficients in our condition.
\[2{x_1} - 36{y_1} - 2 = 0\]
\[ - 36{x_1} + 23{y_1} - 14 = 0\]
Solving above equation we get
\[({x_1},{y_1}) = \left( { - \frac{{11}}{{25}}, - \frac{2}{{25}}} \right)\]
Therefore, the centre of the conic represented by the equation is
\[\left( {\frac{{11}}{{25}}, - \frac{2}{{25}}} \right)\]
Hence option D is correct
Note: We can also find out centre of a conic by the following formula: \[{x_1} = \frac{{hf - bg}}{{ab - {h^2}}},{y_1} = \frac{{gh - af}}{{ab - {h^2}}}\].
Students often make mistakes in these types of problems because these types of problems include many formulas to remember and we should apply the formula to get the desired answer.
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