Q.

Equation of the normal to the hyperbola  3x2−y2=3 at (2,−3)   is

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a

x−2y−8=0

b

3x−2y−12=0

c

x+2y+4=0

d

3x+2y−14=0

answer is A.

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Detailed Solution

a2xx1+b2yy1=a2+b2 x21−y23=1  ⇒a2=1,   b2=3 , (x1,y1)=(2,−3) a2xx1+b2yy1=a2+b2 ⇒1.x2+3.y−3=1+3⇒x2−y=4⇒x−2y=8 ⇒x−2y−8=0
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Equation of the normal to the hyperbola  3x2−y2=3 at (2,−3)   is