Q. 3 D4.4( 5 Votes )

# Find the equation

We need to find the equation of the parabola whose focus is (a, 0) and the vertex is (a’, 0).

We know that the line passing through focus and vertex i.e., the axis is perpendicular to the directrix and vertex is the midpoint of focus and point that lies at the intersection of axis and directrix.

Slope of axis (m1) =

m1 = 0

We know that the products of the slopes of the perpendicular lines is - 1 for non - vertical lines.

Here the slope of the axis is equal to the slope of the x - axis. So, the slope of directrix is equal to the slope of y - axis i.e., ∞.

m2 = ∞

Let us find the point on directrix.

x + a = 2a’ and y = 0

x = 2a’ - a and y = 0

The point on directrix is (2a’ - a, 0).

We know that the equation of the lines passing through (x1, y1) and having slope m is y - y1 = m(x - x1)

y - (0) = ∞(x - (2a’ - a))

x + a - 2a’ = 0

Let us assume P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x1, y1) and (x2, y2) is .

We know that the perpendicular distance from a point (x1, y1) to the line ax + by + c = 0 is .

SP = PM

SP2 = PM2

x2 + y2 - 2ax + a2 = x2 + a2 + 4(a’)2 + 2ax - 4aa’ - 4a’x

y2 - (4a - 4a’)x + a2 - 4(a’)2 + 4aa’ = 0

The equation of the parabola is y2 - (4a - 4a’)x + a2 - 4(a’)2 + 4aa’ = 0.

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