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Question

The line L has intercepts a and b on the coordinate axes. When keeping the origin fixed, the coordinate axes are rotated through a fixed angle, then the same line has intercepts p and q on the rotated axes. Then
(I.I.T. 1990)


A

a2+b2=p2+q2

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B

1a2+1b2=1p2+1q2

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C

a2+b2=p2+q2

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D

1a2+1p2=1b2+1q2

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Solution

The correct option is B

1a2+1b2=1p2+1q2


Suppose the axes are rotated in the anti- clockwise direction through an angle α. The equation of the line L with respect to the old axes is given by xa+yb=1 To find the equation of L with respect to the new axes, we replace x by (x cos α - y sin α) and y by (x sin α + y cos α) so that the equation of L with respect to the new axes is 1a(xcosαysinα)+1b(xsinα+ycosα)=1

Since p, q are the intercepts made by this line on the co-ordinate axes, we have on putting (p, 0) and then (0, q)
1p=1acosα+1bsinα
and 1q=1asinα+1bcosα

Eliminate α. Squaring and adding, we get
1p2+1q2=1a2+1b2

Alternate solution:
xa+yb=1 transforms to xp+xq=1 after rotation of axes. Since origin and the line are fixed only its equation is changed and hence perpendicular from origin is same
11a2+1b2=11p2+1q2
1a2+1b2=1p2+1q2


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