Q. 24 - Calculus

Question

Use Cartesian coordinates to express the equations for the parabolas determined by the conditions specified in Exercises 22–31. 

$\text{ directrix }x=-1,\text{ focus }(2,5)$

Step-by-Step Solution

Answer

The equation is $(y-5{)}^2=6x-3.$

Step 1. Given information.

It is given,

$\text{ directrix }x=-1,\text{ focus }(2,5)$

Step 2. Distance formula.

Now,

$$\begin{gathered} \text{ Let }(x,y)\text{ be any point on the parabola. } \\ \text{ Formula for the distance }=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2} \end{gathered}$$

The distance between the point and the focus is, 

$\text{ Distance }=\sqrt{(2-x{)}^2+(5-y{)}^2}\left[\text{ since }{x}_1=x,y_1=y,x_2=2,y_2=5\right]$

Now the distance between the point and directrix is $|x+1|[sincex=-1]$

Therefore,

$\sqrt{(2-x{)}^2+(5-y{)}^2}=|x+1|$

Step 3. Final answer.

On simplifying the equation,

$$\begin{gathered} {\left(\sqrt{(2-x{)}^2+(5-y{)}^2}\right)}^2=\left(\right|x+1\left|{)}^2 \\ \right(2-x{)}^2+(5-y{)}^2=(x+1{)}^2 \\ \left(4-4x+x^2\right)+\left(25-10y+y^2\right)=\left(x^2+2x+1\right) \\ {y}^2-10y+29-4x=2x+1 \\ {y}^2-10y+29-4x+4x=6x+1 \\ {y}^2-10y+29=6x+1 \\ (y-5{)}^2+4=6x+1 \\ (y-5{)}^2+4-4=6x+1-4 \\ (y-5{)}^2=6x-3 \end{gathered}$$