Question

Let O be the vertex of the parabola \(x^2=4y\) and Q be any point on it. Let the locus of the point P, which divides the line segment OQ internally in the ratio 2:3 be the conic C. Then the equation of the chord of C, which is bisected at the point (1, 2), is:

• A. \(5x-4y+3=0\)
• B. \(x-2y+3=0\)
• C. \(5x-y-3=0\)
• D. \(4x-5y+6=0\)

Hint:

The formula \(T=S_1\) is a very useful shortcut for finding the equation of a chord bisected at a given point for any second-degree conic section (circle, parabola, ellipse, hyperbola).
Remembering the standard substitutions for T is key: \(x^2 \to xx_1\), \(y^2 \to yy_1\), \(x \to \frac{x+x_1}{2}\), \(y \to \frac{y+y_1}{2}\), \(xy \to \frac{xy_1+x_1y}{2}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
First, we need to find the locus of a point P that divides the segment OQ in a given ratio, where O is the vertex of a parabola and Q is a variable point on it. This locus will be a new conic C. Second, we need to find the equation of a chord of this new conic C that is bisected at a given point (1, 2).
Step 2: Key Formula or Approach:
1. Parametric form of parabola: Any point on the parabola \(x^2=4ay\) can be represented as \((2at, at^2)\). For \(x^2=4y\), \(a=1\), so a point Q is \((2t, t^2)\).
2. Section Formula: If a point P(h,k) divides the line segment joining \(A(x_1, y_1)\) and \(B(x_2, y_2)\) in the ratio m:n, then \(P = \left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)\).
3. Equation of chord bisected at a point: The equation of the chord of a conic \(S=0\) which is bisected at the point \((x_1, y_1)\) is given by \(T = S_1\), where T is the expression for the tangent at \((x_1, y_1)\) and \(S_1\) is the value of the conic's expression at \((x_1, y_1)\).
Step 3: Detailed Explanation:
Part 1: Finding the locus of P.
The parabola is \(x^2=4y\). The vertex is O(0,0).
Let any point on the parabola be Q\((2t, t^2)\).
The point P(h,k) divides the segment OQ in the ratio 2:3.
Using the section formula:
\[ h = \frac{2(2t) + 3(0)}{2+3} = \frac{4t}{5} \implies t = \frac{5h}{4} \] \[ k = \frac{2(t^2) + 3(0)}{2+3} = \frac{2t^2}{5} \] To find the locus, we eliminate the parameter \(t\). Substitute \(t\) from the first equation into the second:
\[ k = \frac{2}{5} \left(\frac{5h}{4}\right)^2 = \frac{2}{5} \frac{25h^2}{16} = \frac{5h^2}{8} \] So, \(8k = 5h^2\). The locus of P is the conic C: \(5x^2 = 8y\), or \(5x^2 - 8y = 0\).
Part 2: Finding the equation of the chord.
The conic C is \(S \equiv 5x^2 - 8y = 0\).
The chord is bisected at the point \((x_1, y_1) = (1, 2)\).
We use the formula \(T = S_1\).
To find T, we replace \(x^2 \to xx_1\), \(y \to \frac{y+y_1}{2}\) in the equation of the conic.
\[ T \equiv 5(x \cdot x_1) - 8\left(\frac{y+y_1}{2}\right) = 5x(1) - 4(y+2) = 5x - 4y - 8 \] To find \(S_1\), we substitute \((x_1, y_1)\) into the equation of the conic.
\[ S_1 \equiv 5x_1^2 - 8y_1 = 5(1)^2 - 8(2) = 5 - 16 = -11 \] Now, set \(T = S_1\):
\[ 5x - 4y - 8 = -11 \] \[ 5x - 4y + 3 = 0 \] Step 4: Final Answer:
The equation of the chord of the conic C bisected at (1, 2) is \(5x-4y+3=0\).