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10.2 parabola,ellipse,hyperbola — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS10.2 parabola,ellipse,hyperbola1 Mark
MCQ
The locus of the centroid of the triangle formed by any point $\mathrm{P}$ on the hyperbola $16 \mathrm{x}^{2}-9 \mathrm{y}^{2}+$ $32 x+36 y-164=0$, and its foci is:
  • A
    $9 x^{2}-16 y^{2}+36 x+32 y-36=0$
  • ✓
    $16 x^{2}-9 y^{2}+32 x+36 y-36=0$
  • C
    $16 x^{2}-9 y^{2}+32 x+36 y-144=0$
  • D
    $9 x^{2}-16 y^{2}+36 x+32 y-144=0$
✓

Answer

Correct option: B.
$16 x^{2}-9 y^{2}+32 x+36 y-36=0$
b
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 $(\mathrm{h}, \mathrm{k})$

$\, A(\alpha, \beta)$ be point on hyperbola

So $h=\frac{\alpha-6+4}{3}, k=\frac{\beta+2+2}{3}$

$\Rightarrow \alpha=3 h+2, \beta=3 k-4$

$(\alpha, \beta)$ lies on hyperbola so

$16(3 h+2+1)^{2}-9(3 k-4-2)^{2}=144$

$\Rightarrow 144(h+1)^{2}-81(k-2)^{2}=144$

$\Rightarrow 16\left(h^{2}+2 h+1\right)-9\left(k^{2}-4 k+4\right)=16$

$\Rightarrow 16 x^{2}-9 y^{2}+32 x+36 y-36=0$

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