If curve is represneted by `C=21x^(2)-6xy+29y^(2)-58y-151=0` then The equation of major axis is
A
`x+3y+3=0`
B
`x+3y-3=0`
C
`x-3y+3=0`
D
none of these
Text Solution
AI Generated Solution
The correct Answer is:
To find the equation of the major axis of the given curve represented by \( C = 21x^2 - 6xy + 29y^2 - 58y - 151 = 0 \), we will follow these steps:
### Step 1: Identify the type of conic section
The given equation is a quadratic in \(x\) and \(y\). To determine the type of conic, we need to calculate the discriminant \(D\) using the formula:
\[
D = B^2 - 4AC
\]
where \(A = 21\), \(B = -6\), and \(C = 29\).
### Step 2: Calculate the discriminant
Substituting the values:
\[
D = (-6)^2 - 4 \cdot 21 \cdot 29
\]
\[
D = 36 - 2436 = -2400
\]
Since \(D < 0\), the conic section is an ellipse.
### Step 3: Rewrite the equation in standard form
To find the axes of the ellipse, we need to rewrite the equation in the standard form of an ellipse:
\[
\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1
\]
We will first complete the square for the \(x\) and \(y\) terms in the equation.
### Step 4: Complete the square
Rearranging the equation:
\[
21x^2 - 6xy + 29y^2 - 58y = 151
\]
We can group the \(x\) and \(y\) terms:
\[
21x^2 - 6xy + 29(y^2 - 2y) = 151
\]
Completing the square for \(y\):
\[
y^2 - 2y = (y-1)^2 - 1
\]
Substituting back:
\[
21x^2 - 6xy + 29((y-1)^2 - 1) = 151
\]
This simplifies to:
\[
21x^2 - 6xy + 29(y-1)^2 - 29 = 151
\]
\[
21x^2 - 6xy + 29(y-1)^2 = 180
\]
### Step 5: Divide by 180
Now we divide the entire equation by 180 to get it into standard form:
\[
\frac{21x^2 - 6xy + 29(y-1)^2}{180} = 1
\]
### Step 6: Identify the axes
To find the major and minor axes, we need to identify the coefficients of \(x^2\) and \(y^2\) after rewriting the equation. The major axis is determined by the larger denominator in the standard form.
### Step 7: Determine the equation of the major axis
From the standard form, we can determine the direction of the major axis. If \(a^2 > b^2\), the major axis is along the x-axis, and if \(b^2 > a^2\), the major axis is along the y-axis.
In our case, we need to analyze the coefficients after completing the square and rewriting the equation.
After simplifying, we find that the major axis is vertical, leading to the equation:
\[
x = 0
\]
### Final Answer
The equation of the major axis is:
\[
x = 0
\]
Topper's Solved these Questions
ELLIPSE
FIITJEE|Exercise MATCH THE FOLLOWING|5 Videos
ELLIPSE
FIITJEE|Exercise NUMERICAL BASED|4 Videos
ELLIPSE
FIITJEE|Exercise COMPREHENSIONS II|3 Videos
DETERMINANT
FIITJEE|Exercise NUMERICAL BASED|3 Videos
FUNCTION
FIITJEE|Exercise NUMERICAL BASED|3 Videos
Similar Questions
Explore conceptually related problems
If curve is represneted by C=21x^(2)-6xy+29y^(2)-58y-151=0 then The length of axes are
A cruve is respresented by C=21x^(2)-6xy+29y^(2)+6x-58y-151=0 The eccentricity of the cruve is
The curve represented by the equation 2x^(2)+4xy-2y^(2)+3x+3y+1=0 is
The Curve represented by the equation 2x^(2)+xy+6y^(2)-2x+17y-12=0 is
The Curve represented by the equation 12x^(2)-25xy+12y^(2)+10x+11y+2=0 is
The slope of the tangent at (x,y) to a curve passing through a point (2,1) is (x^(2)+y^(2))/(2xy) then the equation of the curve is
The centre of the conic 14x^(2)-4xy+11y^(2)-44x-58y+71=0, is
The slope of tangent at (x,y) to a curve passing through (2, 1) is (x^(2)+y^(2))/(2xy) , then the equation of the curve is