Question

The centre of the ellipse x²+7y²-14x+28y+49=0 is

• A. (7,0)
• B. (7,-4)
• C. (7,-2)
• D. (-7,4)
• E. (-7,2)

Answer 1

The Correct Option is C

We are given the equation of an ellipse: \[ x^2 + 7y^2 - 14x + 28y + 49 = 0. \] To find the centre of the ellipse, we need to rewrite this equation in the standard form of the ellipse equation: \[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1, \] where \( (h, k) \) is the centre of the ellipse.
Step 1: Group the \( x \)-terms and \( y \)-terms Rearrange the given equation: \[ x^2 - 14x + 7y^2 + 28y = -49. \] Now complete the square for both the \( x \)-terms and \( y \)-terms.
Step 2: Complete the square for the \( x \)-terms Take the coefficient of \( x \), which is \( -14 \), divide it by 2, square it: \[ \left( \frac{-14}{2} \right)^2 = 49. \] Add 49 to both sides: \[ x^2 - 14x + 49 + 7y^2 + 28y = -49 + 49, \] which simplifies to: \[ (x - 7)^2 + 7y^2 + 28y = 0. \]
Step 3: Complete the square for the \( y \)-terms Take the coefficient of \( y \), which is \( 28 \), divide it by 2, square it: \[ \left( \frac{28}{2 \times 7} \right)^2 = 4. \] Add \( 4 \) inside the bracket: \[ (x - 7)^2 + 7(y + 2)^2 = 0. \] Thus, the centre of the ellipse is \( (7, -2) \).

The correct option is (C) : \((7,-2)\)

Answer 2

The given equation is \(x^2 + 7y^2 - 14x + 28y + 49 = 0\).

We complete the square for both x and y terms:

\((x^2 - 14x) + 7(y^2 + 4y) + 49 = 0\)

To complete the square for the x terms, we take half of the coefficient of x (-14) and square it: \((-7)^2 = 49\).

To complete the square for the y terms, we take half of the coefficient of y (4) and square it: \((2)^2 = 4\).

So we have:

\((x^2 - 14x + 49) + 7(y^2 + 4y + 4) + 49 - 49 - 7(4) = 0\)

\((x - 7)^2 + 7(y + 2)^2 - 28 = 0\)

\((x - 7)^2 + 7(y + 2)^2 = 28\)

Dividing by 28, we get:

\(\frac{(x - 7)^2}{28} + \frac{(y + 2)^2}{4} = 1\)

This is the equation of an ellipse in the form \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), where the center is \((h, k)\).

In our case, the center is \((7, -2)\).