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प्रश्न
Find the equation of the tangent to the hyperbola:
3x2 – 4y2 = 12 at the point (4, 3)
उत्तर
The equation of the hyperbola is 3x2 – 4y2 = 12, i.e., `x^2/4 - y^2/3` = 1
Comparing this equation with `x^2/"a"^2 - y^2/"b"^2` = 1, we get
a2 = 4, b2 = 3
The equation of the tangent to the hyperbola
`x^2/"a"^2 - y^2/"b"^2` = 1 at the point P(x1, y1) on it is
`("xx"_1)/"a"^2 - ("yy"_1)/"b"^2` = 1
∴ the equation of the tangent to the given hyperbola at the point (4, 3) is
`("x"(4))/4 - ("y"(3))/3 = 1`
∴ x – y = 1
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