Question

9. The axis of symmetry of the conic
y = ax^2 + bx+c is
1) y=0 2) x = 0 3) b+2ax = 0 4) y=ax
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Answer 1

Given: y = ax² + bx+c

To find:  Axis of symmetry.

Solution:

• As we have given the conic  y = ax² + bx + c, so this equation is of the type X²=4AY.

• So, making the coefficient of x² in the given equation as 1:

            y = a( x² + bx/a) + c

            y - c / a = ( x² + bx/a)

            y - c / a = (x +b/2a)² - b²/4a²

            (x +b/2a)² = y - c/a + b²/4a²

            (x +b/2a)² = (4ay - 4ac + b²) / 4a²

            (x +b/2a)² = 1/4a² [4ay + b² - 4ac]

• Taking 4a common, we get:

            (x +b/2a)² = 4a/4a² [y + ( b² - 4ac / 4a )]

            (x +b/2a)² = 1/a [y + ( b² - 4ac / 4a )]

• So, now X² = 4AY

            so making (x +b/2a)² as zero,

            x = -b/2a

            and making y + ( b² - 4ac / 4a ) as zero,

            y = -b² + 4ac / 4a

• So, x = -b/2a

            2ax = -b

            So the line of symmetry is 2ax + b = 0.

Answer:

    The axis of symmetry of the conic  y = ax² + bx+c is 3) 2ax + b = 0