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Q.
The focus of parabola whose vertex is $(4, 5)$ and whose equation of directrix $x + y+1= 0$ is
Conic Sections
Solution:
Equation of directrix $x + y + 1 = 0$.
Since directrix and axis of parabola are at right angles.
$\therefore $ Equation of axis is $x - y +k=0$ passes through the vertex.
$\therefore k= 1$ (using (4, 5) in $x - y + k = 0)$
$\therefore $ Required equation of axis i s $x - y + 1= 0$
Now vertex is mid point of $NS$
$\therefore \frac{x-1}{2}=4$
$\Rightarrow x=9$ and $\frac{y}{2}=5$
$\Rightarrow y=10$
$\therefore $ Focus is $\left(9,10\right)$Equation of directrix $x + y + 1 = 0$.
Since directrix and axis of parabola are at right angles.
$\therefore $ Equation of axis is $x - y +k=0$ passes through the vertex.
$\therefore k= 1$ (using (4, 5) in $x - y + k = 0)$
$\therefore $ Required equation of axis i s $x - y + 1= 0$
Now vertex is mid point of $NS$
$\therefore \frac{x-1}{2}=4$
$\Rightarrow x=9$ and $\frac{y}{2}=5$
$\Rightarrow y=10$
$\therefore $ Focus is $\left(9,10\right)$
