JEE Main 2022 (Online) 27th June Evening ShiftQuestion 1
If the equation of the parabola, whose vertex is at $(5,4)$ and the directrix is $3x+y-29=0$, is $x^{2}+ay^{2}+bxy+cx+dy+k=0$, then $a+b+c+d+k$ is equal to
Parabola
Solution
Foot of the vertex on the directrix: $(5+3s,4+s)$ with $3(5+3s)+4+s-29=0\Rightarrow s=1$, foot $(8,5)$. Focus $=2V-\text{foot}=(2,3)$. $(x-2)^2+(y-3)^2=\frac{(3x+y-29)^2}{10}$, so $10(x^2+y^2-4x-6y+13)=9x^2+y^2+6xy-174x-58y+841$. $x^2+9y^2-6xy+134x-2y-711=0$, so $a=9,b=-6,c=134,d=-2,k=-711$. $a+b+c+d+k=-576$. Option (D).