Find the equation of parabola :

vertex =(2,1)

directrix : x = y-1

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Find the equation of parabola : 
vertex =(2,1)
directrix : x = y-1

Gursheen kaur. · Accepted Answer

Equation of directrix= x - y+1=0--------(i)

Let the coordinate of focus = (a,b)

$∵$ the axis of parabola is perpendicular to directrix.

So the equation of axis of parabola may be taken as x+y+k=0

$∵$ it passes through (2,1)

⇒ 2+1+k=0

⇒ k=-3

⇒ the equation of axis of parabola is x+y-3=0---------------(ii)

Now, for the point of intersection of directrix and axis of parabola

On solving (i) and (ii), we get,

$y = 2$

put the value of y in (i), we get, $x = 1$

Thus, the point of intersection of directrix and axis of parabola $(x, y) = (1, 2)$

Coordinates of $z = (1, 2)$

$∵$ Vertex A (2,1) is the mid-point of focus and $z = (1, 2)$

$\Rightarrow 2 = \frac{a + 1}{2} a n d 1 = \frac{b + 2}{2} \Rightarrow 4 = a + 1 \Rightarrow 2 = b + 2 \Rightarrow a = 3 \Rightarrow b = 0$

Coordinate of focus = $(3, 0)$

$\Rightarrow \sqrt{{(x - 3)}^{2} + {(y - 0)}^{2}} = | \frac{x - y + 1}{\sqrt{1^{2} + 1^{2}}} | \Rightarrow \sqrt{{(x - 3)}^{2} + y^{2}} = | \frac{x - y + 1}{\sqrt{2}} | s q u a r i n g b o t h s i d e s, w e g e t, \Rightarrow {(x - 3)}^{2} + y^{2} = \frac{{(x - y + 1)}^{2}}{2} o n s o l v i n g t h i s e q u a t i o n, w e g e t, \Rightarrow x^{2} + 9 - 6 x + y^{2} = \frac{x^{2} + y^{2} + 1 - 2 x y - 2 y + 2 x}{2} \Rightarrow 2 x^{2} + 18 - 12 x + 2 y^{2} = x^{2} + y^{2} + 1 - 2 x y - 2 y + 2 x \Rightarrow 2 x^{2} + 18 - 12 x + 2 y^{2} - x^{2} - y^{2} - 1 + 2 x y + 2 y - 2 x = 0 \Rightarrow x^{2} + y^{2} - 14 x + 2 y + 2 x y + 17 = 0$

which is the equation of parabola

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Find the equation of parabola : vertex =(2,1) directrix : x = y-1

Find the equation of parabola :

vertex =(2,1)

directrix : x = y-1