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Miscellaneous Examples · Example 19

Q.A rod ABAB of length 1515 cm rests in between two coordinate axes in such a way that the end point AA lies on xx-axis and end point BB lies on yy-axis. A point P(x,y)P(x, y) is taken on the rod in such a way that AP=6AP = 6 cm. Show that the locus of PP is an ellipse.

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The rod slides with its ends on the axes; point PP divides ABAB in a fixed ratio. Using the section formula and the constraint AB=15AB=15, we eliminate the coordinates of AA and BB to get 4x2+9y2=3244x^2 + 9y^2 = 324, 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 PP is fixed relative to the rod — it is 6 cm from AA and therefore 9 cm from BB. So PP divides ABAB in the ratio AP:PB=6:9=2:3AP:PB = 6:9 = 2:3. This ratio stays constant no matter how the rod slides.

Because AA and BB always lie on the xx-axis and yy-axis respectively, we can write their coordinates as A(a,0)A(a,0) and B(0,b)B(0,b). The distance ABAB is fixed at 15, giving a2+b2=225a^2 + b^2 = 225. The point PP is the point that divides ABAB internally in the ratio 2:32:3. Using the section formula, we express xx and yy in terms of aa and bb, then eliminate aa and bb using the distance constraint.

Let’s work through it.

  1. Set up coordinates. Let A=(a,0)A = (a, 0) on the xx-axis and B=(0,b)B = (0, b) on the yy-axis, where a≥0a \ge 0, b≥0b \ge 0 (the rod is in the first quadrant). The length condition gives:

AB=(a−0)2+(0−b)2=a2+b2=15AB = \sqrt{(a-0)^2 + (0-b)^2} = \sqrt{a^2 + b^2} = 15

So:

a2+b2=225a^2 + b^2 = 225

  1. Locate point PP using the section formula. PP divides ABAB such that AP=6AP = 6 and PB=9PB = 9, so AP:PB=2:3AP:PB = 2:3. For internal division, the coordinates of PP are:

x=3⋅a+2⋅02+3=3a5x = \frac{3 \cdot a + 2 \cdot 0}{2+3} = \frac{3a}{5}

y=3⋅0+2⋅b2+3=2b5y = \frac{3 \cdot 0 + 2 \cdot b}{2+3} = \frac{2b}{5}

Hence:

a=5x3,b=5y2a = \frac{5x}{3}, \quad b = \frac{5y}{2}

  1. Substitute into the length constraint. From a2+b2=225a^2 + b^2 = 225:

(5x3)2+(5y2)2=225\left(\frac{5x}{3}\right)^2 + \left(\frac{5y}{2}\right)^2 = 225

25x29+25y24=225\frac{25x^2}{9} + \frac{25y^2}{4} = 225

  1. Simplify to standard form. Divide through by 25:

x29+y24=9\frac{x^2}{9} + \frac{y^2}{4} = 9

Multiply through by 36 (LCM of 9 and 4) to clear denominators:

4x2+9y2=3244x^2 + 9y^2 = 324

Or, in standard ellipse form: …

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