Q.A rod of length cm rests in between two coordinate axes in such a way that the end point lies on -axis and end point lies on -axis. A point is taken on the rod in such a way that cm. Show that the locus of is an ellipse.
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Start your 14-day free trial to unlock the full solution →The rod slides with its ends on the axes; point divides in a fixed ratio. Using the section formula and the constraint , we eliminate the coordinates of and to get , which is an ellipse.
The key insight is that the rod’s ends are constrained to the axes, but the rod itself can take any orientation. Point is fixed relative to the rod — it is 6 cm from and therefore 9 cm from . So divides in the ratio . This ratio stays constant no matter how the rod slides.
Because and always lie on the -axis and -axis respectively, we can write their coordinates as and . The distance is fixed at 15, giving . The point is the point that divides internally in the ratio . Using the section formula, we express and in terms of and , then eliminate and using the distance constraint.
Let’s work through it.
- Set up coordinates. Let on the -axis and on the -axis, where , (the rod is in the first quadrant). The length condition gives:
So:
- Locate point using the section formula. divides such that and , so . For internal division, the coordinates of are:
Hence:
- Substitute into the length constraint. From :
- Simplify to standard form. Divide through by 25:
Multiply through by 36 (LCM of 9 and 4) to clear denominators:
Or, in standard ellipse form: …
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