An Interesting Conic Problem!
The equation of a conic is,
(x^2) +116x + 4(y^2) + 2y + 4xy + 259 = 0. Find out what kind of conic it is. Then, Let, the ways we can permutate the letters of the name of This conic is X. Let, the directix of this conic is in the form ax+by+c=0. Let, the focus of This conic is ( p,q ).
Then,
Find out the value of ( X+(a)^2+(b)^2+(c)^2+(p)^2+(q)^2 )-172
The answer is 6702.
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1 solution
We know, the general equation of the conic is, ax^2 + 2hxy + by^2 +2gx + 2fy + c = 0 . And, if ab=h^2, then the conic is parabola. Comparing the given equation with this general equation, we get a=1, b=4 & h=2. Thus, 1.4= 2^2 means, ab=h^2. And so, the given equation is of the conic Parabola. So, now we can permutate the letters of the name "Parabola" in (8!/3!)=6720 ways. So, X=6720. Then, Rearranging the equation of This conic, we can rewrite it as following :- ((x+2y)^2) + 116x + 2y + 259 = 0 ................(1)
Now, we know, if a parabola has directix, ax+by+c=0 & focus (p,q), Then it's general equation (I've said GENERAL, not IDEAL) is as following :-
((bx-ay)^2) -2x( (a^2+b^2)p+ac) - 2y( (a^2+b^2)q+bc) + (a^2+b^2)(p^2+q^2) - (c^2)=0. Now, comparing the (1) no. equation with this general equation and after doing some general calculation, we get, a=(-2), b=(1), c= (9) & p=(-8), q=(-2).
So, (X+(a)^2+(b)^2+(c)^2+(p)^2+(q)^2)-172= (6720+4+1+81+64+4-172) = 6702 is the answer.