Question

The equation of the parabola whose focus is $(3,-4)$ and directrix$6x-7y+5=0$, is

• A. ${(7x+6y)}^2–570x+750y+2100=0$
• B. ${(7x+6y)}^2+570x-750y+2100=0$
• C. ${(7x-6y)}^2–570x+750y+2100=0$
• D. ${(7x-6y)}^2+570x-750y+2100=0$

Answer 1

The correct option is A

${(7x+6y)}^2–570x+750y+2100=0$

Step 1: Using the definition of a parabola , find the relation between focal distances

Let $F(3,-4)$ be the focus of the parabola.

Let $M:6x-7y+5=0$ be the directrix of the parabola

Any point $P(x,y)$ lying on the parabola is equidistant from the focus and the directrix

$\Rightarrow$ $FP=MP$

$\Rightarrow$$F{P}^2=M{P}^2$

Step 2: Apply distance formula

$FP=\sqrt{{\left(x-3\right)}^2+{\left(y+4\right)}^2}$

$\Rightarrow$ $F{P}^2={\left(x-3\right)}^2+{\left(y+4\right)}^2$ $...\left(i\right)$

Step 3: Apply formula for perpendicular distance between point and line

$MP=\frac{\left|\left(6x-7y+5\right)\right|}{\sqrt{6^2+{(-7)}^2}}$

$\Rightarrow$ $M{P}^2=\frac{{\left(6x-7y+5\right)}^2}{6^2+{(-7)}^2}$ $...\left(ii\right)$

From $\left(i\right),\left(ii\right)$

${\left(x-3\right)}^2+{\left(y+4\right)}^2=\frac{{(6x-7y+5)}^2}{85}$

$\Rightarrow$ $85x^2-510x+765+85y^2+680y+1360=36x^2+49y^2+25+60x-70y-84xy$

$\Rightarrow$ $49x^2+36y^2+84xy-570x+750y+2100=0$

$\Rightarrow$ ${(7x+6y)}^2-570x+750y+2100=0$

Hence option $\left(A\right)$ is the correct answer.