Question

# The equation of the parabola whose vertex is (a, 0) and the directrix has the equation x + y = 3a, is (a) x2 + y2 + 2xy + 6ax + 10ay + 7a2 = 0 (b) x2 − 2xy + y2 + 6ax + 10ay − 7a2 = 0 (c) x2 − 2xy + y2 − 6ax + 10ay − 7a2 = 0 (d) none of these

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Solution

## (b) x2 − 2xy + y2 + 6ax + 10ay − 7a2 = 0 Given: The vertex is at (a, 0) and the directrix is the line x + y = 3a. The slope of the line perpendicular to x + y = 3a is 1. The axis of the parabola is perpendicular to the directrix and passes through the vertex. ∴ Equation of the axis of the parabola = $y-0=1\left(x-a\right)$ (1) Intersection point of the directrix and the axis is the intersection point of (1) and x + y = 3a. Let the intersection point be K. Therefore, the coordinates of K are $\left(2a,a\right)$. The vertex is the mid-point of the segment joining K and the focus (h, k). ∴ $a=\frac{2a+h}{2},0=\frac{a+k}{2}\phantom{\rule{0ex}{0ex}}h=0,k=-a$ Let P (x, y) be any point on the parabola whose focus is S (h, k) and the directrix is x + y = 3a. Draw PM perpendicular to x + y = 3a. Then, we have: $SP=PM\phantom{\rule{0ex}{0ex}}⇒S{P}^{2}=P{M}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-0\right)}^{2}+{\left(y+a\right)}^{2}={\left(\frac{x+y-3a}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{\left(y+a\right)}^{2}={\left(\frac{x+y-3a}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+2{y}^{2}+2{a}^{2}+4ay={x}^{2}+{y}^{2}+9{a}^{2}+2xy-6ax-6ay\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-7{a}^{2}+10ay+6ax-2xy=0$

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