If the equation of the parabola, whose vertex is at (5, 4) and the directrix is $$3x + y - 29 = 0$$, is $${x^2} + a{y^2} + bxy + cx + dy + k = 0$$, then $$a + b + c + d + k$$ is equal to :
JEE · Math · previous-year question
- A.575
- B.$$-$$575
- C.576
- D.$$-$$576correct
Answer
D. $$-$$576
Explanation
Given vertex is (5, 4) and directrix 3x + y $$-$$ 29 = 0 Let foot of perpendicular of (5, 4) on directrix is (x1, y1) $${{{x_1} - 5} \over 3} = {{{y_1} - 4} \over 1} = {{ - ( - 10)} \over {10}}$$ $$\therefore$$ $$({x_1},\,{y_1}) \equiv (8,\,5)$$ So, focus of parabola will be $$S = (2,3)$$ Let P(x, y) be any point on parabola, then $${(x - 2)^2} + {(y - 3)^2} = {{{{(3x + y - 29)}^2}} \over {10}}$$ $$ \Rightarrow 10({x^2} + {y^2} - 4x - 6y + 13) = 9{x^2} + {y^2} + 841 + 6xy - 58y - 174x$$ $$ \Rightarrow {x^2} + 9{y^2} - 6xy + 134x - 2y - 711 = 0$$ and given parabola $${x^2} + a{y^2} + bxy + cx + dy + k = 0$$ $$\therefore$$ a = 9, b = $$-$$6, c = 134, d = $$-$$2, k = $$-$$711 $$\therefore$$ $$a + b + c + d + k = - 576$$
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