Question

The equation of the parabola whose vertex is (a,0) and the directrix has the equation x+y=3a,is

• A. $x^2+y^2+2xy+6ax+10ay+7a^2=0$
• B. $x^2-2xy+y^2+6ax+10ay-7a^2=0$
• C. $x^2-2xy+y^2-6ax+10ay-7a^2=0$
• D. none of these

Answer 1

The correct option is B

$x^2-2xy+y^2+6ax+10ay-7a^2=0$

Given: The vertex is at (a,0) and the directrix is the lline x+y=3a

The slope of the line perpendicular to x+y is 1

The axis of the parabola is perpendicular to the directrix and passes through the vertex.

$\therefore \text{Equation of the axis of the parabola=y-0=1(x-a)}...\left(1\right)$

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 (2a,a)

The vertex is the mid-point of the segment joining K and the focus (h,k).

$\therefore a=\frac{2a+h}{2}=0,0=\frac{a+k}{2}$

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

$\Rightarrow S{P}^2=P{M}^2$

$\Rightarrow (x-0{)}^2+(y+a{)}^2={\left(\frac{x+y-3a}{\sqrt{2}}\right)}^2$

$\Rightarrow x^2+(y+a{)}^2={\left(\frac{x+y-3a}{\sqrt{2}}\right)}^2$

$\Rightarrow 2x^2+2y^2+2a^2+4ay$

$=x^2+y^2+9a^2+2xy-6ax-6ay$

$\Rightarrow x^2+y^2-7a^2+10ay+6ax-2xy=0$