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Question

Let C be the centre of the hyperbola x2a2−y2b2=1. The tangent at any point P on this hyperbola meets the straight lines bx−ay=0 and bx+ay=0 in the points Q and R respectively. Then CQ.CR is equal to:

A
a2+b2
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B
|a2b2|
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C
1a2+1b2
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D
a2b2a2+b2
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Solution

The correct option is A a2+b2
Given hyperbola : x2a2y2b2=1
equation of tangent at P(x1,y1) is
T : xx1a2yy1b2=1 ....... (1)
(wherex12a2y12b2=1) ......... (2)
Point of intersection of tangent T and lines
of y=bax xx1a2y1ab=1
x=(a2+y1ab)1x1
&y=(ab+y1)1x1
Q : ((a2+y1ab)1x1,(ab+y1)1x1)
of y=bax x=(a2y1ab)1x1y=(aby1)1x1
R : ((a2y1ab)1x1,(aby1)1x1)
To find ¯¯¯¯¯¯¯¯¯CQ.¯¯¯¯¯¯¯¯CR=[(a2+y1ab)1x1]2+[(ab+y1)1x1]2.[(a2y1ab)1x1]2+[(aby1)1x1]2
= 1x12[a4y12a2b2]+1x12[a2b2y12]2+1x14[2(a3by12ab)]
= 1x14[x12(a4y12a2b2y12+a2b2)+2(a3by12ab)]
Now put (x1,y1)as(asecα,btanα) and they do give
a4+2a2b2+b2
=a2+b2

996470_1035158_ans_1ce398a7cb2d41dc8e0c11a5849c63e9.png

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