Question

Find the equation of the parabola whose focus is $S(3,5)$ and vertex is $A(1,3)$.

• A. $\begin{array}{c}\\ \Rightarrow \left|\right|={\left(x+y\right)}^2=2\left[{\left(x-3\right)}^2+{\left(y-5\right)}^2\right]\end{array}$
  
• B. $\begin{array}{c}\\ \Rightarrow \left|\right|={\left(x+y\right)}^2=2\left[{\left(x-6\right)}^2+{\left(y-6\right)}^2\right]\end{array}$
  
• C. $\begin{array}{c}\\ \Rightarrow \left|\right|={\left(x+y\right)}^2=2\left[{\left(x-11\right)}^2+{\left(y-11\right)}^2\right]\end{array}$
  
• D. $\begin{array}{c}\\ \Rightarrow \left|\right|={\left(x+y\right)}^2=2\left[{\left(x-7\right)}^2+{\left(y-7\right)}^2\right]\end{array}$

Answer 1

The correct option is A $\begin{array}{c}\\ \Rightarrow \left|\right|={\left(x+y\right)}^2=2\left[{\left(x-3\right)}^2+{\left(y-5\right)}^2\right]\end{array}$


Slope of axis$=\frac{5-3}{3-1}=\frac{2}{2}=1$
Slope of directrix$=-1$
equation of tangent at vertex $A$
$\begin{array}{c}Pt(1,3),m=-1\\ \Rightarrow y-3=-1\left(x-1\right)\\ \Rightarrow x+y-4=0\end{array}$
equation of directrix
$\begin{array}{c}x+y=\lambda \\ a=SA\\ =\sqrt{4+4}=2\sqrt{2}\end{array}$
$A$ is midpoint of $PS$
$\begin{array}{c}\frac{n+3}{2}=1,\frac{k+5}{2}=3\\ \Rightarrow n=-1,k=1\end{array}$
$(-1,1)$ lies on directrix
$-1+1=\lambda =0$
equation of diectrix: $L:y+x=0$
$\begin{array}{c}QO=QS\\ \Rightarrow \left|\frac{l+m}{\sqrt{2}}\right|=\sqrt{{\left(l-3\right)}^2+{\left(m-5\right)}^2}\\ \Rightarrow {\left(l+m\right)}^2=2\left[{\left(l-3\right)}^2+{\left(m-5\right)}^2\right]\\ \Rightarrow {\left(x+y\right)}^2=2\left[{\left(x-3\right)}^2+{\left(y-5\right)}^2\right]\end{array}$