Solved on Nov 18, 2023

Find the equation of a conic section without rotating the coordinate system. $3x^2 - 5xy + 9y^2 - 17 = 0$

STEP 1

Assumptions1. The given equation is a second degree equation in two variables x and y. . The equation is in the form of $ax^ +hxy + by^ + c =0$ , where a, h, b, and c are constants.
3. We need to identify the type of the equation without applying a rotation of axes.

STEP 2

First, we need to identify the type of the given equation. The type of the equation is determined by the values of a, h, and b.The given equation is $x^2 -5xy +9y^2 -17 =0$ .Comparing this with the general form $ax^2 +2hxy + by^2 + c =0$ , we get $a =$ , $2h = -5$ , and $b =9$ .So, $h = -5/2$ .

STEP 3

Now, we need to check the sign of $ab - h^2$ to identify the type of the equation.The formula is $ab - h^2$ .

STEP 4

Substitute the values of a, b, and h into the formula.
$ab - h^2 =3 \\times9 - (-/2)^2$

STEP 5

Calculate the value of $ab - h^2$ .
$ab - h^2 =27 -.25 =20.75$

STEP 6

Since $ab - h^2 >0$ and $a >0$ , the given equation is an ellipse.
The equation of the ellipse is $3x^2 -5xy +9y^2 -17 =0$ .

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Find the equation of a conic section without rotating the coordinate system. $3x^2 - 5xy + 9y^2 - 17 = 0$

STEP 1

Assumptions1. The given equation is a second degree equation in two variables x and y. . The equation is in the form of $ax^ +hxy + by^ + c =0$ , where a, h, b, and c are constants.
3. We need to identify the type of the equation without applying a rotation of axes.

STEP 2

First, we need to identify the type of the given equation. The type of the equation is determined by the values of a, h, and b.The given equation is $x^2 -5xy +9y^2 -17 =0$ .Comparing this with the general form $ax^2 +2hxy + by^2 + c =0$ , we get $a =$ , $2h = -5$ , and $b =9$ .So, $h = -5/2$ .

STEP 3

Now, we need to check the sign of $ab - h^2$ to identify the type of the equation.The formula is $ab - h^2$ .

STEP 4

Substitute the values of a, b, and h into the formula.
$ab - h^2 =3 \\times9 - (-/2)^2$

STEP 5

Calculate the value of $ab - h^2$ .
$ab - h^2 =27 -.25 =20.75$

STEP 6

Since $ab - h^2 >0$ and $a >0$ , the given equation is an ellipse.
The equation of the ellipse is $3x^2 -5xy +9y^2 -17 =0$ .

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Find the equation of a conic section without rotating the coordinate system. 3x2−5xy+9y2−17=03x^2 - 5xy + 9y^2 - 17 = 03x2−5xy+9y2−17=0

STEP 1

Assumptions1. The given equation is a second degree equation in two variables x and y. . The equation is in the form of ax+hxy+by+c=0ax^ +hxy + by^ + c =0ax+hxy+by+c=0, where a, h, b, and c are constants.3. We need to identify the type of the equation without applying a rotation of axes.

STEP 2

STEP 3

Now, we need to check the sign of ab−h2ab - h^2ab−h2 to identify the type of the equation.The formula is ab−h2ab - h^2ab−h2.

STEP 4

Substitute the values of a, b, and h into the formula.ab−h2=3times9−(−/2)2ab - h^2 =3 \\times9 - (-/2)^2ab−h2=3times9−(−/2)2

STEP 5

Calculate the value of ab−h2ab - h^2ab−h2.ab−h2=27−.25=20.75ab - h^2 =27 -.25 =20.75ab−h2=27−.25=20.75

STEP 6

Since ab−h2>0ab - h^2 >0ab−h2>0 and a>0a >0a>0, the given equation is an ellipse.The equation of the ellipse is 3x2−5xy+9y2−17=03x^2 -5xy +9y^2 -17 =03x2−5xy+9y2−17=0.

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Find the equation of a conic section without rotating the coordinate system. $3x^2 - 5xy + 9y^2 - 17 = 0$

Assumptions1. The given equation is a second degree equation in two variables x and y. . The equation is in the form of $ax^ +hxy + by^ + c =0$ , where a, h, b, and c are constants.
3. We need to identify the type of the equation without applying a rotation of axes.

Now, we need to check the sign of $ab - h^2$ to identify the type of the equation.The formula is $ab - h^2$ .

Substitute the values of a, b, and h into the formula.
$ab - h^2 =3 \\times9 - (-/2)^2$

Calculate the value of $ab - h^2$ .
$ab - h^2 =27 -.25 =20.75$

Since $ab - h^2 >0$ and $a >0$ , the given equation is an ellipse.
The equation of the ellipse is $3x^2 -5xy +9y^2 -17 =0$ .