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Question

From point (2,2) tangents are drawn to the hyperbola x216y29=1 then the point of contacts lie in

A
I and II are quadrants
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B
I and IV quadrants
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C
I and III quadrants
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D
III and IV quadrants
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Solution

The correct option is D III and IV quadrants
Let m1&m2 be slopes of tangents to x216y29=1 from (2,2)
m1&m2 are roots of the equation -
(x1a2)m22x1y1m+(y21+b2)=0
(216)m28m+(4+9)=0
14m28m+13=0
m1=0.7194
m2=1.2908
Equation of tangents are -
(y2)=0.7194(x2);(y2)=1.2908(x2)
y=0.7194x+0.5162;y=1.2908x+4.5816
Putting these values in equation of hyperbola one by one. We find that points of contact lie on 3rd and 4th quadrants.
Hence, the answer is III and IV quadrants.



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