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

Given the equation of the parabola whose vertex is $(5,4)$(5,4) and equation of directrix is $3x+y-29=0$3x+y−29=0 is ${x}^{2}+a{y}^{2}+bxy+cx+dy+e=0$x2+ay2+bxy+cx+dy+e=0

Let focus be $(\alpha ,\beta )$(α,β)

Let equation of $ZV$ZV which is perpendicular to $3x+y-29=0$3x+y−29=0 be $x-3y+k=0$x−3y+k=0 and it passes through point $(5,4)$(5,4). So, $k=7$k=7 and equation of $ZV$ZV become $x-3y+7=0$x−3y+7=0

Now the intersection of 3x+y-29=0&x-3y+7=0 will be $Z=(8,5)$Z=(8,5)

Now plotting the diagram we get,

So, foot of perpendicular from $(5,4)$(5,4) on $3x+y-29=0$3x+y−29=0 is $(8,5)$(8,5)

Now using the midpoint formula we will find the focus of parabola$\Rightarrow \frac{\alpha +8}{2}=5,\frac{\beta +5}{2}=4\Rightarrow (\alpha ,\beta )=(2,3)$⇒2α+8​=5,2β+5​=4⇒(α,β)=(2,3)

$\therefore$∴ Focus is $(2,3)$(2,3) and directrix is $3x+y-29=0$3x+y−29=0

Applying ${(PS)}^{2}={(PM)}^{2}$(PS)2=(PM)2, we get

$${(x-2)}^{2}+{(y-3)}^{2}=\frac{{(3x+y-29)}^{2}}{10}$$(x−2)2+(y−3)2=10(3x+y−29)2​

$$\Rightarrow {x}^{2}+9{y}^{2}-6xy+134x-2y-711=0$$⇒x2+9y2−6xy+134x−2y−711=0

Comparing with${x}^{2}+a{y}^{2}+bxy+cx+dy+e=0$x2+ay2+bxy+cx+dy+e=0 , we get $(a+b+c+d+e)=(9-6+134-2-711)=-576$(a+b+c+d+e)=(9−6+134−2−711)=−576