The eccentricity of a hyperbola whose transverse axis is along x− axis and passes through the point (6,4) is √53. The equation of tangent to this hyperbola at (6,4) is
A
x+y−2=0
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B
x−y−2=0
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C
x−y+2=0
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D
x+y+2=0
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Solution
The correct option is Bx−y−2=0 e=√53⇒e2=1+b2a2 ⇒2a2=3b2⋯(i)
Assuming the equation of the hyperbola be, x2a2−y2b2=1
Putting the point (6,4) in the equation of hyperbola, 36a2−16b2=1
Using equation (i), ⇒a2=12⇒b2=8
x212−y28=1⇒x6−yy′4=0
Putting the point (6,4) dydx=1
Equation of tangent is, (y−4)=1(x−6) ⇒x−y−2=0