Let \(O\) be the vertex of the parabola \(y^{2}=16x\). The locus of the centroid of \(\triangle OPA\), when point \(P\) lies on the parabola and point \(A\) lies on the \(x\)-axis such that \(\angle OPA = 90^\circ\), is:

Question

Let \( y = x \) be the equation of a chord of the circle \( C_1 \) (in the closed half-plane \( x \ge 0 \)) of diameter 10 passing through the origin. Let \( C_2 \) be another circle described on the given chord as diameter. If the equation of the chord of the circle \( C_2 \), which passes through the point \( (2, 3) \) and is farthest from the center of \( C_2 \), is \( x + ay + b = 0 \), then \( b \) is equal to:

Answer 1

To find the locus of the centroid of the triangle \(\triangle OPA\), we start by considering the given parabola \(y^2 = 16x\), which has its vertex at the origin \(O(0,0)\). The standard form \(y^2 = 4ax\) indicates that here \(a = 4\).

Let \(P\) be a point on the parabola. Hence, the coordinates of \(P\) are \( (x_1, y_1) \) such that \( y_1^2 = 16x_1 \).

Let \(A\) be a point on the \(x\)-axis. So, coordinates of \(A\) can be assumed to be \( (x_2, 0) \).

It is given that \(\angle OPA = 90^\circ\). The dot product of vectors \(\overrightarrow{OP}\) and \(\overrightarrow{PA}\) is zero due to the orthogonal condition. Therefore:

\[ \overrightarrow{OP} \cdot \overrightarrow{PA} = 0 \]

The vectors are \(\overrightarrow{OP} = (x_1, y_1)\) and \(\overrightarrow{PA} = (x_2 - x_1, 0 - y_1)\). So, the condition becomes:

\[ x_1(x_2 - x_1) + y_1(0 - y_1) = 0 \]

\[ x_1 x_2 - x_1^2 - y_1^2 = 0 \]

Since \(y_1^2 = 16x_1\), substitute \(16x_1\) in place of \(y_1^2\):

\[ x_1 x_2 - x_1^2 - 16x_1 = 0 \]

Rearrange to find \(x_2\):

\[ x_2 = x_1 + 16 \]

The coordinates of the centroid \(G\) of \(\triangle OPA\) are given by:

\[ G = \left(\frac{0 + x_1 + x_2}{3}, \frac{0 + y_1 + 0}{3}\right) \]

Substitute \(x_2 = x_1 + 16\) into the centroid coordinates:

\[ G = \left(\frac{0 + x_1 + (x_1 + 16)}{3}, \frac{y_1}{3}\right) = \left(\frac{2x_1 + 16}{3}, \frac{y_1}{3}\right) \]

Replace \(x_1\) using the parabola equation \(y_1^2 = 16x_1\):

\[ y_1^2 = 16 \cdot \frac{3}{2}(x - \frac{16}{3}) = 24x - 128 \]

Substituting \((24x - 128)\) in place of \(x_1\), we derive the condition for \(y_1\) :

\[ \frac{y_1^2}{3} = 8x - \frac{128}{3} \Rightarrow 9y_1^2 = 8(3x - 16) \]

Thus, the locus of the centroid of \(\triangle OPA\) is:

\[ 9y^2 = 8(3x - 16) \]

This confirms that the correct answer is: \( 9y^2 = 8(3x - 16) \)

Answer 2