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

अतिपरवलय, \(16 x ^{2}-9 y ^{2}+32 x +36 y -164=0\) पर किसी बिंदु \(P\) तथा इसकी नाभियों से बने त्रिभुज के केन्द्रक का बिन्दुपथ है

  1. A \(9 x^{2}-16 y^{2}+36 x+32 y-36=0\)
  2. B \(16 x^{2}-9 y^{2}+32 x+36 y-36=0\)
  3. C \(16 x^{2}-9 y^{2}+32 x+36 y-144=0\)
  4. D \(9 x^{2}-16 y^{2}+36 x+32 y-144=0\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(16 x^{2}-9 y^{2}+32 x+36 y-36=0\)

Step-by-step Solution

Detailed explanation

Given hyperbola is \(16(x+1)^{2}-9(y-2)^{2}=164+16-36=144\) \(\Rightarrow \frac{(x+1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1\) \(\text { Eccentricity, } e=\sqrt{1+\frac{16}{9}}=\frac{5}{3}\) \(\Rightarrow \text { foci are }(4,2) \text { and }(-6,2)\) Let the centroic be…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app