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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

यदि परवलय का समीकरण, जिसका शीर्ष बिन्दु \((5,4)\) पर तथा नियता \(3 x + y -29=0\) हो, \(x^2+a y^2+b x y+c x+d y+k=0\) है, तो \(a + b + c + d + k\) बराबर है

  1. A \(575\)
  2. B \(-575\)
  3. C \(576\)
  4. D \(-576\)
Verified Solution

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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