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प्रश्न
Answer the following:
Find the equation of the tangent to the hyperbola 7x2 − 3y2 = 51 at (−3, −2)
उत्तर
The equation of the hyperbola is 7x2 − 3y2 = 51
i.e. `x^2/((51/7)) - y^2/17` = 1
Comparing with `x^2/"a"^2 - y^2/"b"^2` = 1, we get,
a2 = `51/7`, b2 = 17
The equation of the tangent to `x^2/"a"^2 - y^2/"b"^2` = 1 at the point (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 (–3, –2) is
`(-3x)/((51/7)) + (2y)/17` = 1
∴ `(-7x)/17 + (2y)/17` = 1
∴ 7x – 2y + 17 = 0.
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