JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
A stair-case of length \(l\) rests against a vertical wall and a floor of a room. Let \(P\) be a point on the stair-case, nearer to its end on the wall, that divides its length in the ratio \(1 : 2\). If the staircase begins to slide on the floor, then the locus of \(P\) is
- A an ellipse of eccentricity \(\frac{1}{2}\)
- B an ellipse of eccentncity \(\frac{{\sqrt 3 }}{2}\)
- C a circle of radius \(\frac{1}{2}\)
- D a circle of radius \(\frac{{\sqrt 3 }}{2}l\)
Answer & Solution
Correct Answer
(B) an ellipse of eccentncity \(\frac{{\sqrt 3 }}{2}\)
Step-by-step Solution
Detailed explanation
Let point \(A\left( {a,0} \right)\)is on \(x\)-axis and \(B\left( {0,b} \right)\)is on \(y\)-axis. Let \(P(h,k)\) divides \(AB\) in the ration \(1:2\) So, by section formula \(h = \frac{{2\left( 0 \right) + 1\left( a \right)}}{{1 + 2}} = \frac{a}{3}\)…
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