Question:
If \(O\) is the origin and \(P\) is a point moving on the straight line \(lx + my + n = 0 \; (n \neq 0)\). If \(Q\) is a point on the segment \(OP\) such that \(OP \cdot OQ = k^2\), where \(k \neq 0\), then the locus of \(Q\) is
If \(O\) is the origin and \(P\) is a point moving on the straight line \(lx + my + n = 0 \; (n \neq 0)\). If \(Q\) is a point on the segment \(OP\) such that \(OP \cdot OQ = k^2\), where \(k \neq 0\), then the locus of \(Q\) is
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Whenever a point divides a segment joining the origin and another moving point, use proportional coordinates. Then convert the given geometric condition into algebraic form and substitute into the given curve or line equation.
Updated On: Jun 17, 2026
- \(n(x^2+y^2)=k^2(lx+my)\)
- \(k^2(x^2+y^2)=n(lx+my)\)
- \(n(x^2-y^2)=k^2(lx-my)\)
- \(n(x^2+y^2)=k^2(ly-mx)\)
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The Correct Option is A
Solution and Explanation
Concept:
This problem is based on the geometrical interpretation of section formula and locus formation.
A point \(Q\) lies on the segment joining the origin \(O\) and another moving point \(P\).
The relation between distances \(OP\) and \(OQ\) helps us express coordinates of \(P\) in terms of coordinates of \(Q\).
Substituting these coordinates into the equation of the moving line gives the required locus.
Step 1: Assume coordinates of the moving point \(P\).
Let \[ P(\alpha,\beta) \] be any point lying on the line \[ l\alpha + m\beta + n =0. \] Since \(Q\) lies on the segment \(OP\), let \[ Q(x,y). \] Because \(Q\) lies on the same line joining the origin and \(P\), the vectors are proportional. Hence, \[ Q=\lambda P. \] Therefore, \[ x=\lambda \alpha,\qquad y=\lambda \beta. \]
Step 2: Use the given condition \(OP\cdot OQ=k^2\).
Now, \[ OP=\sqrt{\alpha^2+\beta^2} \] and \[ OQ=\sqrt{x^2+y^2}. \] Since \[ x=\lambda \alpha,\qquad y=\lambda \beta, \] we obtain \[ x^2+y^2=\lambda^2(\alpha^2+\beta^2). \] Hence, \[ OQ=\lambda OP. \] Given, \[ OP\cdot OQ=k^2. \] Substituting \(OQ=\lambda OP\), \[ OP(\lambda OP)=k^2 \] which gives \[ \lambda(OP)^2=k^2. \] But \[ (OP)^2=\alpha^2+\beta^2. \] Therefore, \[ \lambda(\alpha^2+\beta^2)=k^2. \] Using \[ x=\lambda\alpha,\qquad y=\lambda\beta, \] we get \[ \alpha=\frac{x}{\lambda},\qquad \beta=\frac{y}{\lambda}. \] Substituting, \[ \lambda\left(\frac{x^2+y^2}{\lambda^2}\right)=k^2 \] which simplifies to \[ \lambda=\frac{x^2+y^2}{k^2}. \]
Step 3: Substitute into the equation of the line.
Since \(P(\alpha,\beta)\) lies on \[ l\alpha+m\beta+n=0, \] we substitute \[ \alpha=\frac{x}{\lambda},\qquad \beta=\frac{y}{\lambda}. \] Thus, \[ l\left(\frac{x}{\lambda}\right)+m\left(\frac{y}{\lambda}\right)+n=0. \] Multiplying throughout by \(\lambda\), \[ lx+my+n\lambda=0. \] Now substitute \[ \lambda=\frac{x^2+y^2}{k^2}. \] Hence, \[ lx+my+n\left(\frac{x^2+y^2}{k^2}\right)=0. \] Multiplying by \(k^2\), \[ k^2(lx+my)+n(x^2+y^2)=0. \] Rearranging, \[ n(x^2+y^2)=k^2(lx+my). \] Hence the required locus is \[ \boxed{n(x^2+y^2)=k^2(lx+my)}. \]
Step 1: Assume coordinates of the moving point \(P\).
Let \[ P(\alpha,\beta) \] be any point lying on the line \[ l\alpha + m\beta + n =0. \] Since \(Q\) lies on the segment \(OP\), let \[ Q(x,y). \] Because \(Q\) lies on the same line joining the origin and \(P\), the vectors are proportional. Hence, \[ Q=\lambda P. \] Therefore, \[ x=\lambda \alpha,\qquad y=\lambda \beta. \]
Step 2: Use the given condition \(OP\cdot OQ=k^2\).
Now, \[ OP=\sqrt{\alpha^2+\beta^2} \] and \[ OQ=\sqrt{x^2+y^2}. \] Since \[ x=\lambda \alpha,\qquad y=\lambda \beta, \] we obtain \[ x^2+y^2=\lambda^2(\alpha^2+\beta^2). \] Hence, \[ OQ=\lambda OP. \] Given, \[ OP\cdot OQ=k^2. \] Substituting \(OQ=\lambda OP\), \[ OP(\lambda OP)=k^2 \] which gives \[ \lambda(OP)^2=k^2. \] But \[ (OP)^2=\alpha^2+\beta^2. \] Therefore, \[ \lambda(\alpha^2+\beta^2)=k^2. \] Using \[ x=\lambda\alpha,\qquad y=\lambda\beta, \] we get \[ \alpha=\frac{x}{\lambda},\qquad \beta=\frac{y}{\lambda}. \] Substituting, \[ \lambda\left(\frac{x^2+y^2}{\lambda^2}\right)=k^2 \] which simplifies to \[ \lambda=\frac{x^2+y^2}{k^2}. \]
Step 3: Substitute into the equation of the line.
Since \(P(\alpha,\beta)\) lies on \[ l\alpha+m\beta+n=0, \] we substitute \[ \alpha=\frac{x}{\lambda},\qquad \beta=\frac{y}{\lambda}. \] Thus, \[ l\left(\frac{x}{\lambda}\right)+m\left(\frac{y}{\lambda}\right)+n=0. \] Multiplying throughout by \(\lambda\), \[ lx+my+n\lambda=0. \] Now substitute \[ \lambda=\frac{x^2+y^2}{k^2}. \] Hence, \[ lx+my+n\left(\frac{x^2+y^2}{k^2}\right)=0. \] Multiplying by \(k^2\), \[ k^2(lx+my)+n(x^2+y^2)=0. \] Rearranging, \[ n(x^2+y^2)=k^2(lx+my). \] Hence the required locus is \[ \boxed{n(x^2+y^2)=k^2(lx+my)}. \]
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