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Question: How do you determine circle, parabola, ellipse, or hyperbola from equation \(4{{x}^{2}}+9{{y}^{2}}-1...

How do you determine circle, parabola, ellipse, or hyperbola from equation 4x2+9y2−16x+18y−11=04{{x}^{2}}+9{{y}^{2}}-16x+18y-11=0 ?

Explanation

Solution

These questions should be generally solved by converting the equation given in the question to standard form. After arriving at the standard equation, you can compare it with the standard equations of hyperbola, ellipse, parabola and circle and check if it matches any of the equations of the figures. If it matches, then you get your final answer.

Complete step by step answer:
According to the problem, we are asked to determine if the following equation is a circle, parabola, ellipse, or hyperbola from equation 4x2+9y2−16x+18y−11=04{{x}^{2}}+9{{y}^{2}}-16x+18y-11=0 ?
We take the equation as equation 1.
4x2+9y2−16x+18y−11=04{{x}^{2}}+9{{y}^{2}}-16x+18y-11=0--- ( 1 )
Now, we convert this equation 1 to standard form.
⇒4x2+9y2−16x+18y−11=0\Rightarrow 4{{x}^{2}}+9{{y}^{2}}-16x+18y-11=0
⇒4x2−16x+9y2+18y−11=0\Rightarrow 4{{x}^{2}}-16x+9{{y}^{2}}+18y-11=0
⇒4x2−16x+16+9y2+18y+9−11−16−9=0\Rightarrow 4{{x}^{2}}-16x+16+9{{y}^{2}}+18y+9-11-16-9=0
⇒4(x2−4x+4)+9(y2+2y+1)−11−16−9=0\Rightarrow 4\left( {{x}^{2}}-4x+4 \right)+9\left( {{y}^{2}}+2y+1 \right)-11-16-9=0
⇒4(x−2)2+9(y+1)2=36\Rightarrow 4{{\left( x-2 \right)}^{2}}+9{{\left( y+1 \right)}^{2}}=36
⇒(x−2)29+(y+1)24=1\Rightarrow \dfrac{{{\left( x-2 \right)}^{2}}}{9}+\dfrac{{{\left( y+1 \right)}^{2}}}{4}=1 ---- (2)
Equation 2 is the standard equation. From here, we can tell that this is an ellipse. To be more precise, it is a horizontal ellipse.
So, we have found whether the given equation is a circle, parabola or ellipse, or a hyperbola. The equation is an ellipse.
Therefore, the solution of the given equation 4x2+9y2−16x+18y−11=04{{x}^{2}}+9{{y}^{2}}-16x+18y-11=0 is ellipse.

Note:
While solving these questions, you should be careful while forming the whole squares. You should remember to subtract or add the values on both the sides. The easier way to solve these questions is to check if the coefficients of x2{{x}^{2}} and y2{{y}^{2}} are the same. If they are the same, then it is a circle or else it is an ellipse. If you have y2{{y}^{2}}or x2{{x}^{2}} negative, then it is a hyperbola. If you have only either of x2{{x}^{2}} or y2{{y}^{2}}, then it is parabola.