Home
Class 12
MATHS
The equation of the major axis of 25x^(2...

The equation of the major axis of `25x^(2) + 16y^(2)-100x - 96y - 156 = 0` is

A

x-4=0

B

x-2=0

C

y-3=0

D

y-6=0

Text Solution

AI Generated Solution

The correct Answer is:
To find the equation of the major axis of the given ellipse \(25x^2 + 16y^2 - 100x - 96y - 156 = 0\), we will follow these steps: ### Step 1: Rearrange the equation Start by rearranging the equation to group the \(x\) and \(y\) terms together: \[ 25x^2 - 100x + 16y^2 - 96y - 156 = 0 \] ### Step 2: Complete the square for \(x\) terms Factor out the coefficient of \(x^2\) from the \(x\) terms: \[ 25(x^2 - 4x) + 16y^2 - 96y - 156 = 0 \] Now, complete the square for \(x^2 - 4x\): \[ x^2 - 4x = (x - 2)^2 - 4 \] So, substituting this back, we get: \[ 25((x - 2)^2 - 4) + 16y^2 - 96y - 156 = 0 \] This simplifies to: \[ 25(x - 2)^2 - 100 + 16y^2 - 96y - 156 = 0 \] ### Step 3: Complete the square for \(y\) terms Now, factor out the coefficient of \(y^2\) from the \(y\) terms: \[ 25(x - 2)^2 + 16(y^2 - 6y) - 256 = 0 \] Complete the square for \(y^2 - 6y\): \[ y^2 - 6y = (y - 3)^2 - 9 \] Substituting this back, we have: \[ 25(x - 2)^2 + 16((y - 3)^2 - 9) - 256 = 0 \] This simplifies to: \[ 25(x - 2)^2 + 16(y - 3)^2 - 144 = 0 \] ### Step 4: Rearranging the equation Rearranging gives: \[ 25(x - 2)^2 + 16(y - 3)^2 = 144 \] ### Step 5: Divide by 144 Now, divide the entire equation by 144 to get it into standard form: \[ \frac{25(x - 2)^2}{144} + \frac{16(y - 3)^2}{144} = 1 \] This simplifies to: \[ \frac{(x - 2)^2}{\frac{144}{25}} + \frac{(y - 3)^2}{\frac{144}{16}} = 1 \] Which can be written as: \[ \frac{(x - 2)^2}{5.76} + \frac{(y - 3)^2}{9} = 1 \] ### Step 6: Identify the major and minor axes From the standard form \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), we see: - \(a^2 = 5.76\) (which means \(a = \sqrt{5.76} = 2.4\)) - \(b^2 = 9\) (which means \(b = 3\)) Since \(b > a\), the major axis is vertical. ### Step 7: Write the equation of the major axis The center of the ellipse is at \((h, k) = (2, 3)\). The equation of the major axis, which is vertical, is given by: \[ x = h = 2 \] Thus, the equation of the major axis is: \[ \boxed{x = 2} \]
Promotional Banner

Similar Questions

Explore conceptually related problems

The length of the transverse axis of the hyperbola 9x^(2)-16y^(2)-18x -32y - 151 = 0 is

The equation of the auxiliary circle of x^(2)/(16)-y^(2)/(25)=1 is

The equations of the latus recta of the hyperbola 9x^(2) -16y^(2) -18x -32y -151 =0 are

The foci of the conic 25x^(2) +16y^(2)-150 x=175 are :

Obtain the parametric equation of the circle represented by x^(2) + y^(2) + 6x + 8y- 96= 0

The length of the conjugate axis of the hyperbola 9x^(2) -16y^(2) - 18x -64y + 89 =0 is a) 8 b) 6 c) 4 d) 5

The equations of the asymptotes of the conic. x^(2) + 24xy - 6y^(2) + 28x + 36y + 16 = 0 are

The length of the major axis of the ellipse 3x^2+2y^2-6x8y-1=0 is

The length of the latus rectum of the hyperbola 9x^(2) -16y^(2) +72x -32y- 16 =0 is

Find the equation of the radical axis of the following circles. x^2+y^2-2x-4y-1=0 x^2+y^2-4x-6y+5=0