Question

The locus of the centroid of the triangle formed by any point P on the hyperbola $16x^2 - 9y^2 + 32x + 36y - 164 = 0$, and its foci is :

• A. $9x^2 - 16y^2 + 36x + 32y - 36 = 0$
• B. $16x^2 - 9y^2 + 32x + 36y - 144 = 0$
• C. $9x^2 - 16y^2 + 36x + 32y - 144 = 0$
• D. $16x^2 - 9y^2 + 32x + 36y - 36 = 0$

Hint:

If the centroid of a triangle with two fixed vertices $A$ and $B$ and one moving vertex $P$ (lying on curve $C$) is required, the locus will always be a scaled and shifted version of the original curve $C$. Specifically, the linear dimensions are scaled by $1/3$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To find the locus, we first express the equation of the hyperbola in standard form to identify its center and foci.
We then use the centroid formula for a triangle with vertices \(P(x_1, y_1)\), \(F_1(x_2, y_2)\), and \(F_2(x_3, y_3)\).
Finally, we substitute the coordinates of \(P\) (expressed in terms of centroid coordinates) into the hyperbola equation.
Step 3: Detailed Explanation:
Standardizing the hyperbola equation:
\(16(x^2 + 2x) - 9(y^2 - 4y) = 164\)
\(16(x+1)^2 - 16 - 9(y-2)^2 + 36 = 164\)
\(16(x+1)^2 - 9(y-2)^2 = 144\)
Divide by 144: \(\frac{(x+1)^2}{9} - \frac{(y-2)^2}{16} = 1\).
Center is \(C(-1, 2)\). Here \(a^2=9, b^2=16\).
Eccentricity \(e = \sqrt{1 + b^2/a^2} = \sqrt{1 + 16/9} = 5/3\).
Distance from center to focus is \(ae = 3 \cdot (5/3) = 5\).
Foci (\(F_1, F_2\)) are shifted from \(C\): \((-1 \pm 5, 2)\), which are \((4, 2)\) and \((-6, 2)\).
Let \(P(x_0, y_0)\) be a point on the hyperbola.
Let \((h, k)\) be the centroid of \(\triangle PF_1F_2\):
\(h = \frac{x_0 + 4 + (-6)}{3} \Rightarrow 3h = x_0 - 2 \Rightarrow x_0 = 3h + 2\).
\(k = \frac{y_0 + 2 + 2}{3} \Rightarrow 3k = y_0 + 4 \Rightarrow y_0 = 3k - 4\).
Substitute \(x_0\) and \(y_0\) into the standardized hyperbola equation:
\[ \frac{(3h + 2 + 1)^2}{9} - \frac{(3k - 4 - 2)^2}{16} = 1 \]
\[ \frac{(3h + 3)^2}{9} - \frac{(3k - 6)^2}{16} = 1 \Rightarrow \frac{9(h + 1)^2}{9} - \frac{9(k - 2)^2}{16} = 1 \]
\[ (h + 1)^2 - \frac{9(k - 2)^2}{16} = 1 \Rightarrow 16(h+1)^2 - 9(k-2)^2 = 16 \]
Expand the equation:
\(16(h^2 + 2h + 1) - 9(k^2 - 4k + 4) = 16\)
\(16h^2 + 32h + 16 - 9k^2 + 36k - 36 = 16\)
\(16h^2 - 9k^2 + 32h + 36k - 36 = 0\).
Replacing \((h, k)\) with \((x, y)\) gives the locus.
Step 4: Final Answer:
The locus is \(16x^2 - 9y^2 + 32x + 36y - 36 = 0\).