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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
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