find the equation to the tangent to the hyperbola x2/16-y2/9=1 which are parallel to the line y2=3x-4
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Find the equation of the tangent to the curve \( x^2+3y=3\), which is parallel to the line y−4x+5=0
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Equation of the tangent at (x1,y1) where slope is m is given by y−y1=m(x−x1)
Step 1:
Given curve is x2+3y=3
Let y=−x2+33
On differentiating with respect to x
∴dydx=13[−2x]
=−23x
Step 2:
Since the tangent is parallel to the line y−4x+5=0
Then slopes should be equal
Slope of the given line is 4
−2x3=4
x=−4×32
=−122
=−6
Step 3:
∴y=−(−6)2+33
=−36+33
=−333
=−11
Hence the points of contact are (−6,−11)
Step 4:
Equation of the tangent at (x1,y1) where slope is m is given by y−y1=m(x−x1)
(i.e)[y−(−11)]=4(x−(−6))
y+11=4x+6
⇒4x−y=11−6
⇒4x−y=5
Hence 4x−y=5 is the required equation.