Question

equation of parabola when point of intersection of directrix and axis is (0,4) and focus is at (6,4)​

Answer 1

Answer:

hi I am a $tupid and I have a big but

Answer 2

Given,

Focus $\[ = S\left( {6,4} \right)\]$

Point of intersection of directrix and axis $\[ = A\left( {0,4} \right)\]$

Solution,

Slope of axis,

$\[\begin{array}{l}m = \frac{{4 - 4}}{{6 - 0}}\\ \Rightarrow m = 0\end{array}\]$

Know that axis and directrix both are perpendicular to each other.

Slope of directrix,

$\[m' = \frac{{ - 1}}{0}\]$

Equation of directrix,

$\[\begin{array}{l}y - 4 = \frac{{ - 1}}{0}\left( {x - 0} \right)\\ \Rightarrow x = 0\end{array}\]$

Calculate the equation of parabola,

$\[\begin{array}{l}PS = PM\\ \Rightarrow \left( {\sqrt {{{\left( {x - 6} \right)}^2} + {{\left( {y - 4} \right)}^2}} } \right) = \frac{x}{{\sqrt {{1^2}} }}\end{array}\]$

Squaring on both side,

$\[\begin{array}{l} \Rightarrow {\left( {x - 6} \right)^2} + {\left( {y - 4} \right)^2} = {x^2}\\ \Rightarrow {x^2} - 12x + 36 + {y^2} - 8y + 16 = {x^2}\\ \Rightarrow {y^2} - 12x - 8y + 52 = 0\end{array}\]$

Hence the equation of parabola is $\[{y^2} - 12x - 8y + 52 = 0\]$