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

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Solution

(i) Let P(x,y) be any point on the parabola whose focus is S(3,0) and the directrix 3x+4y=1.Draw PM perpendicular from P(x,y) on the directrix 3x+4y=1

Then by definition SP=PM

$\Rightarrow S P^{2} = P M^{2} \Rightarrow (x - 3)^{2} + (y - 0)^{2} = {(\frac{3 x + 4 y - 1}{\sqrt{(3)^{2} + (4)^{2}}})}^{2} \Rightarrow x^{2} + 9 - 6 x + y^{2} = \frac{(3 x + 4 y - 1)^{2}}{\sqrt{(25)^{2}}} \Rightarrow x^{2} - 6 x + y^{2} + 9 = \frac{(3 x + 4 y - 1)^{2}}{25} \Rightarrow 25 (x^{2} - 6 x + y^{2} + 9) = (3 x + 4 y - 1)^{2} \Rightarrow 25 x^{2} - 150 x + 25 y^{2} + 225 = (3 x)^{3} + (4 y)^{2} + (- 1)^{2} + 2 \times 3 x \times 4 y + 2 \times 4 y \times (- 1) + 2 \times (- 1) \times 3 x \Rightarrow 25 x^{2} - 150 x + 25 y^{2} + 225 = 9 x^{2} + 16 y^{2} + 1 + 24 x y - 8 y - 6 x \Rightarrow 25 x^{2} - 9 x^{2} + 25 y^{2} - 16 y^{2} - 150 x + 6 x + 8 y - 24 x y + 225 - 1 = 0 \Rightarrow 16 x^{2} + 9 y^{2} - 144 x + 8 y - 24 x y + 224 = 0 \Rightarrow 16 x^{2} + 9 y^{2} - 24 x y - 144 x + 8 y + 224 = 0$

This is the equation of the required parabola.

(ii) Let P(x,y) be any point on the parabola whose focus is S(1,1) and the directrix x+y+1=0.Draw PM perpendicular from P(x,y) on the directrix x+y+1=0

Then by definition SP=PM

$\Rightarrow S P^{2} = P M^{2} \Rightarrow (x - 1)^{2} + (y - 1)^{2} = {(\frac{x + y + 1}{\sqrt{(1)^{2} + (1)^{2}}})}^{2} \Rightarrow x^{2} + 1 - 2 x + y^{2} + 1 - 2 y = (\frac{(x + y + 1)^{2}}{\sqrt{2}}) \Rightarrow x^{2} + y^{2} - 2 x - 2 y + 2 = \frac{(x + y + 1)^{2}}{2} \Rightarrow 2 (x^{2} + y^{2} - 2 x - 2 y + 2) = x^{2} + y^{2} + 1 + 2 x y + 2 y + 2 x \Rightarrow 2 x^{2} + 2 y^{2} - 4 x - 4 y + 4 = x^{2} + y^{2} + 1 + 2 x y + 2 y + 2 x \Rightarrow 2 x^{2} - x^{2} + 2 y^{2} - y^{2} - 2 x y - 4 x - 2 x - 4 y - 2 y + 4 - 1 = 0 \Rightarrow x^{2} + y^{2} - 2 x y - 6 x - 6 y + 3 = 0 This is the equation of the required parabola.$

(iii) Let P(x,y) be any point on the parabola whose focus is S(0,0) and the directrix 2x-y-1=0.

Draw PM perpendicular from P(x,y) on the directrix 2x-y-1=0.

Then by definition SP=PM

$\Rightarrow S P^{2} = P M^{2} \Rightarrow (x - 0)^{2} + (y - 0)^{2} = {(\frac{2 x - y - 1}{\sqrt{(2)^{2} + (- 1)^{2}}})}^{2} \Rightarrow x^{2} + y^{2} = (\frac{(2 x - y - 1)^{2}}{\sqrt{(5)^{2})}}) \Rightarrow 5 (x^{2} + y^{2}) = (2 x - y - 1)^{2} \Rightarrow 5 x^{2} + 5 y^{2} = (2 x)^{2} + (- y)^{2} + (- 1)^{2} + 2 \times 2 x (- y) + 2 (- y) \times (- 1) + 2 \times (- 1) \times 2 x \Rightarrow 5 x^{2} + 5 y^{2} = 4 x^{2} + y^{2} + 1 - 4 x y + 2 y - 4 x \Rightarrow 5 x^{2} - 4 x^{2} + 5 y^{2} - y^{2} + 4 x y + 4 x - 2 y - 1 = 0 \Rightarrow x^{2} + 4 y^{2} + 4 x y + 4 x - 2 y - 1 = 0 This is the equation of the required parabola.$

(iv) Let P(x,y) be any point on the parabola whose focus is S(2,3) and the directrix x-4y+3=0.Draw PM perpendicular from P(x,y) on the directrix x-4y+3=0.

Then by definition SP=PM

$\Rightarrow S P^{2} = P M^{2} \Rightarrow (x - 2)^{2} + (y - 3)^{2} = {(\frac{x - 4 y + 3}{\sqrt{(1)^{2} + (- 4)^{2}}})}^{2} \Rightarrow x^{2} + 4 - 4 x + y^{2} + 9 - 6 y = (\frac{(x - 4 y + 3)^{2}}{\sqrt{(17)^{2})}}) \Rightarrow x^{2} + y^{2} - 4 x - 6 y + 4 + 9 = \frac{(x - 4 y + 3)^{2}}{17} \Rightarrow 17 (x^{2} + y^{2} - 4 x - 6 y + 13) = (x - 4 y + 3)^{2} \Rightarrow 17 x^{2} + 17 y^{2} - 68 x - 102 y + 221 = x^{2} + (- 4 y)^{2} + 3^{2} + 2 \times 3 \times x \Rightarrow 17 x^{2} + 17 y^{2} - 68 x - 102 y + 221 = x^{2} + 16 y^{2} + 9 - 8 x y - 24 y + 6 x \Rightarrow 17 x^{2} - x^{2} + 17 y^{2} - 16 y^{2} + 8 x y - 68 x - 6 x - 102 y + 24 y + 221 - 9 = 0 \Rightarrow 16 x^{2} + y^{2} + 8 x y - 74 x - 78 y + 212 = 0 This is the equation of the required parabola.$

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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

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