JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If the equation of the parabola, whose vertex is at \((5,4)\) and the directrix is \(3 x+y-29=0\), is \(x^{2}+a y^{2}+b x y+c x+d y+k=0\) then \(a + b + c + d + k\) is equal to
- A \(575\)
- B \(-575\)
- C \(576\)
- D \(-576\)
Answer & Solution
Correct Answer
(D) \(-576\)
Step-by-step Solution
Detailed explanation
Vertex \((5,4)\) Directrix : \(3 x+y-29=0\) Co-ordinates of B (foot of directrix) \(\frac{x-5}{3}=\frac{y-4}{1}=-\left(\frac{15+4-29}{10}\right)=1\) \(x=8, y=5\) \(S =(2,3)\) (focus) Equation of parabola \(PS = PM\) so equation is \(x^{2}+9 y^{2}-6 x y+134 x-2 y-711=0\)…
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