KVPY SX SB Exam 2016 Question Paper

Question : 84 of 120

Marks: +1, -0

A wall is inclined to the floor at an angle of 135º. A ladder of length l is resting on the wall. As the ladder slides down, its mid-point traces an arc of an ellipse. Then the area of the ellipse is

${πl^2}/4$
${πl^2}$
${4πl^2}$
${2πl^2}$

Validate

Solution:

Mid point (h, k)

=({x-x_1}/2,x_1/2)

Now

(x + x_1)^2 + x_1^2 = l^2

2h + 4k = x + x_1, 2k = x,

So required locus is

4(h + 2k)^2 + 4k^2 = l^2

h^2 + 5k^2 + 4hk=l^2/4

x^2 + 5y^2 + 4xy=l^2/4

Whose area is

{πl^2}/4

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