Find the equation of parabola whose
Focus is $(1, 1)$ and the directrix is $x + y + 1 = 0$

Open in App

Solution

focus = (1,1)
$e q^{n}$ directrix $= x + y + 1 = 0$
Let A(x,y) be the pt. on the parabola
& AM be the length of $⊥$ from A to
directrix
Then SA = AM
$\Rightarrow (S A)^{2} = (A M)^{2}$
$\Rightarrow (x - 1)^{2} + (y - 1^{2}) = \frac{(x + y + 1)^{2}}{(\sqrt{1^{2} + 1^{2}})^{2}}$
$\Rightarrow x^{2} + 1 - 2 x + y^{2} + 1 - 2 y = \frac{x^{2} + y^{2} + 1 + 2 x y + 2 y + 2 x}{2}$
$\Rightarrow 2 x^{2} + 2 - 4 x + 2 y^{2} + 2 - 4 y = x^{2} + y^{2} + 1 + 2 x y + 2 x + 2 y$
$\Rightarrow x^{2} + y^{2} - 6 x - 6 y - 2 x y + 3 = 0$
Which is the required equation of parabola.

Suggest Corrections

Similar questions

(- 1, 1)

x + y + 1 = 0

(1, 1)

x - y + 1 = 0

Find the equation of the parabola whose:

(i) focus is (3,0) and the directrix is 3x+4y=1

(ii) focus is (1,1) and the directrix is x+y+1=0

(iii) focus is (0,0) and the directrix is 2x-y-1=0

(iv) focus is (2,3) and the directrix is x-4y+3=0

(1, - 1)

x + y + 7 = 0

(- 3, 0)

x + 5 = 0

Join BYJU'S Learning Program