Question

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 2x − y − 1 = 0
(iv) focus is (2, 3) and the directrix x − 4y + 3 = 0.

Answer 1

(i) Let P (x, y) be any point on the parabola whose focus is S (3, 0) and the directrix is 3x + 4y = 1.
Draw PM perpendicular to 3x + 4y = 1.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-3\right)}^2+{\left(y-0\right)}^2={\left(\frac{3x+4y-1}{\sqrt{9+16}}\right)}^2 \\ \Rightarrow {\left(x-3\right)}^2+y^2={\left(\frac{3x+4y-1}{5}\right)}^2 \\ \Rightarrow 25\left\{{\left(x-3\right)}^2+y^2\right\}={\left(3x+4y-1\right)}^2 \\ \Rightarrow \left(25x^2-150x+25y^2+225\right)=9x^2+16y^2+1+24xy-8y-6x \\ \Rightarrow 16x^2+9y^2-24xy-144x+8y+224=0 \end{gathered}$$

(ii) Let P (x, y) be any point on the parabola whose focus is S (1, 1) and the directrix is x + y + 1 = 0.
Draw PM perpendicular to x + y + 1 = 0.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-1\right)}^2+{\left(y-1\right)}^2={\left|\frac{x+y+1}{\sqrt{1+1}}\right|}^2 \\ \Rightarrow {\left(x-1\right)}^2+{\left(y-1\right)}^2={\left(\frac{x+y+1}{\sqrt{2}}\right)}^2 \\ \Rightarrow 2\left(x^2+1-2x+y^2+1-2y\right)=x^2+y^2+1+2xy+2y+2x \\ \Rightarrow \left(2x^2+2-4x+2y^2+2-4y\right)=x^2+y^2+1+2xy+2y+2x \\ \Rightarrow x^2+y^2-2xy-6x-6y+3=0 \end{gathered}$$

(iii) Let P (x, y) be any point on the parabola whose focus is S (0, 0) and the directrix is 2x − y − 1 = 0.
Draw PM perpendicular to 2x − y − 1 = 0.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-0\right)}^2+{\left(y-0\right)}^2={\left|\frac{2x-y-1}{\sqrt{4+1}}\right|}^2 \\ \Rightarrow x^2+y^2={\left(\frac{2x-y-1}{\sqrt{5}}\right)}^2 \\ \Rightarrow 5x^2+5y^2=4x^2+y^2+1-4xy+2y-4x \\ \Rightarrow x^2+4y^2+4xy-2y+4x-1=0 \end{gathered}$$

(iv) Let P (x, y) be any point on the parabola whose focus is S (2, 3) and the directrix is x − 4y + 3 = 0.
Draw PM perpendicular to x − 4y + 3 = 0.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-2\right)}^2+{\left(y-3\right)}^2={\left|\frac{x-4y+3}{\sqrt{1+16}}\right|}^2 \\ \Rightarrow {\left(x-2\right)}^2+{\left(y-3\right)}^2={\left(\frac{x-4y+3}{\sqrt{17}}\right)}^2 \\ \Rightarrow 17\left(x^2+4-4x+y^2-6y+9\right)=x^2+16y^2+9-8xy-24y+6x \\ \Rightarrow \left(17x^2-68x-102y+17y^2+13\times 17\right)=x^2+16y^2+9-8xy-24y+6x \\ \Rightarrow 16x^2+y^2+8xy-74x-78y+212=0 \end{gathered}$$