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Question
Find the equation of the tangent to the hyperbola:
`x^2/16 - y^2/9` = 1 at the point in a first quadratures whose ordinate is 3
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Solution
Given equation of the hyperbola is `x^2/16 - y^2/9` = 1.
Comparing this equation with `x^2/"a"^2 - y^2/"b"^2` = 1, we get,
a2 = 16 and b2 = 9
Let P(x1, 3) be the point on the hyperbola in the first quadrant at which the tangent is drawn.
Substituting x = x1 and y = 3 in equation of hyperbola, we get
`x_1^2/16 - 3^2/9` = 1
∴ `x_1^2/16 - 1` = 1
∴ `x_1^2/16` = 2
∴ `x_1^2` = 32
∴ x1 = `± 4sqrt(2)`
Since P lies in the first quadrant,
P ≡ `(4sqrt(2), 3)`
Equation of the tangent to the hyperbola
`x^2/"a"^2 - y^2/"b"^2` = 1 at (x1, y1) is
`("xx"_1)/"a"^2 - ("yy"_1)/"b"^2` = 1
Equation of the tangent at P`(4sqrt(2),3)` is
`(4sqrt(2)x)/16 - (3y)/9` = 1
∴ `sqrt(2)/4x - y/3` = 1
∴ `3sqrt(2)x - 4y` = 12
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