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

To solve this problem, we start by understanding the geometry involved. We are given a parabola with the equation \(x^2 = 4y\), and a point \( O \) which serves as the vertex of this parabola. Any point \( Q \) on this parabola can be represented as \( Q(x_1, \frac{x_1^2}{4}) \).

The point \( P \) divides the line segment \( OQ \) in the ratio 2:3 internally. The coordinates of \( P \) can be calculated using the section formula:

The section formula for a point dividing a line segment internally in the ratio \( m:n \) is:

\(\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)\)  

Here, for point \( P \), \( O \) is (0, 0) and \( Q \) is \( (x_1, \frac{x_1^2}{4}) \), and the ratio is 2:3:

The coordinates of \( P \) are:

\(x = \frac{2*x_1 + 3*0}{2+3} = \frac{2x_1}{5}\)

\(y = \frac{2*\frac{x_1^2}{4} + 3*0}{2+3} = \frac{x_1^2}{10}\)

The locus of point \( P \) as \( Q \) moves along the parabola will be a conic. From the above expressions, eliminate \( x_1 \) by expressing it in terms of \( x \) and \( y \). We have:

From \( x = \frac{2x_1}{5} \), therefore \( x_1 = \frac{5x}{2} \)

Substitute \( x_1 \) in the expression for \( y \), we get:

\(y = \frac{\left(\frac{5x}{2}\right)^2}{10} = \frac{25x^2}{40} = \frac{5x^2}{8}\)

This implies the equation of the locus is:

\(5x^2 = 8y\)

This is the equation of the conic \( C \). Now, we need to find the equation of the chord of this conic which is bisected at the point (1, 2).

The equation of the chord of a parabola (or conic) that is bisected by a point (h, k) is given by:

\(T = S_1\)

Where \( T \) is the equation obtained by replacing \( x^2 \) by \( xx_1 \), \( y^2 \) by \( yy_1 \), and so on in the expression of the curve. \( S_1 \) is the value of the equation at the midpoint (h, k).

Substitute \( (h, k) = (1, 2) \) into \( 5x^2 - 8y = 0 \), and the chord equation becomes:

\(5xh - 8k = 5h^2 - 8k\)

This simplifies to:

\(5x - 8 \cdot 2 = 5 \cdot 1^2 - 8 \cdot 2\)

\(5x - 16y = -11\)

Re-arranging gives:

\(5x - 4y + 3 = 0\)

Thus, the equation of the chord is: Finally, the answer is \(\(5x - 4y + 3 = 0\)\), which matches the given option.