Question

# Find the eqns of the tangent to curve 3x²-y²=8 which passes through the point (4/3,0).

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Solution

## 3x2 - y2 = 8 (i) Differentiating wrt x y' = 3x/y Now since the point passes through (4/3, 0) (0 - y) = 3x/y (4/3 - x) -y2 = 4x - 3x2 (ii) From (i) and (ii) 8 - 3x2 = 4x - 3x2 x = 2 Put the values in any eq. y = (+-) 2 Now solve normally, (y - 2) = 3(x - 2) and 2nd eq. (y + 2) = -3(x - 2) you will get the answer (HOPE YOU LIKED MY ANSWER 😊😊😊)

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