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Question: The vertex of a parabola is \(\left( a,0 \right)\) and the directrix is \(x+y=3a.\) The equation of ...

The vertex of a parabola is (a,0)\left( a,0 \right) and the directrix is x+y=3a.x+y=3a. The equation of the parabola is
(a)x2−2xy+y2+6ax+10ay−7a2=0\left( a \right) {{x}^{2}}-2xy+{{y}^{2}}+6ax+10ay-7{{a}^{2}}=0
(b)x2+2xy+y2+6ax+10ay+2a2=0\left( b \right) {{x}^{2}}+2xy+{{y}^{2}}+6ax+10ay+2{{a}^{2}}=0
(c)x2+2xy+y2+6ax+10ay=2a2\left( c \right) {{x}^{2}}+2xy+{{y}^{2}}+6ax+10ay=2{{a}^{2}}
(d)\left( d \right) None of these

Explanation

Solution

We will find the slope of the line perpendicular to the directrix of the parabola. Then using that, we will find the equation of the axis of the parabola. We will find the intersection point of the directrix and the axis of the parabola. Using this equation, we can find the equation of the given parabola.

Complete step by step solution:
We are asked to find the equation of the parabola whose vertex is (a,0)\left( a,0 \right) and directrix is x+y=3a.x+y=3a.
First, we need to find the slope of the line perpendicular to the directrix for the axis of parabola is perpendicular to the directrix.
Now we know that the perpendicular lines have slopes that are the negative of the reciprocal of each other.
So, we will get the slope of the line x+y=3a,x+y=3a, as −1.-1.
Therefore, we can say that the slope of the line perpendicular to the directrix is 1.1.
That means, the slope of the axis of the parabola is 1.1.
Since the axis of parabola passes through the vertex (a,0)\left( a,0 \right) and perpendicular to the directrix, we can write the equation of the axis of the parabola as y−0=1(x−a).y-0=1\left( x-a \right).
So, we will get y=x−a⇒x−y=a.y=x-a\Rightarrow x-y=a.
We need to find the intersection point of the directrix and the axis.
Now, we solve the equations of the directrix and the axis to find the intersection point.
We will get 2x=4a2x=4a when we add the equations. Then, from this, we will get x=2a.x=2a.
When we apply this value in any of the equations, we will get y=a.y=a.
So, the point of intersection is (2a,a).\left( 2a,a \right).
We know that the vertex is the midpoint of the segment joining the point of intersection of the directrix and the axis and the focus (h,k).\left( h,k \right).
Therefore, we will get a=2a+h2,0=a+k2.a=\dfrac{2a+h}{2},0=\dfrac{a+k}{2}.
From these equations, we can find the coordinate of the focus. We will get h=0h=0 and k=−a.k=-a.
We know that the equation of a parabola with focus (h,k)\left( h,k \right) and directrix y=mx+by=mx+b where mm is the slope is (x−h)2+(y−k)2=(y−mx−b)2m2+1.{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}=\dfrac{{{\left( y-mx-b \right)}^{2}}}{{{m}^{2}}+1}.
No, we will get (x−0)2+(y+a)2=(x+y−3a2)2{{\left( x-0 \right)}^{2}}+{{\left( y+a \right)}^{2}}={{\left( \dfrac{x+y-3a}{\sqrt{2}} \right)}^{2}}
So, we will simplify it to get x2+(y+a)2=(x+y−3a)22.{{x}^{2}}+{{\left( y+a \right)}^{2}}=\dfrac{{{\left( x+y-3a \right)}^{2}}}{2}.
We will get x2+y2+2ay+a2=x2+y2+2xy+9a2−6ax−6ay2.{{x}^{2}}+{{y}^{2}}+2ay+{{a}^{2}}=\dfrac{{{x}^{2}}+{{y}^{2}}+2xy+9{{a}^{2}}-6ax-6ay}{2}.
Now, we will get 2x2+2y2+4ay+2a2=x2+y2+2xy+9a2−6ax−6ay.2{{x}^{2}}+2{{y}^{2}}+4ay+2{{a}^{2}}={{x}^{2}}+{{y}^{2}}+2xy+9{{a}^{2}}-6ax-6ay.
Therefore, we will get x2+y2−2xy−7a2+6ax+10ay=0.{{x}^{2}}+{{y}^{2}}-2xy-7{{a}^{2}}+6ax+10ay=0.
Hence the equation of the parabola is x2+y2−2xy−7a2+6ax+10ay=0.{{x}^{2}}+{{y}^{2}}-2xy-7{{a}^{2}}+6ax+10ay=0.

So, the correct answer is “Option A”.

Note: We know that the parabola is a set of points in a plane which are in equal distance from a fixed point called focus and a fixed line called directrix. The directrix never touches the points of the parabola.