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Question
Answer the following:
Find the equations of the tangents to the hyperbola 3x2 − y2 = 48 which are perpendicular to the line x + 2y − 7 = 0
Solution
The equations of tangents to the hyperbola `x^2/"a"^2 - y^2/"b"^2` = 1 in terms of slope m are
y = `"m"x ± sqrt("a"^2"m"^2 - "b"^2)` ...(1)
The equation of the hyperbola is 3x2 – y2 = 48
i.e., `x^2/16 - y^2/48` = 1
Comparing this with `x^2/"a"^2 - y^2/"b"^2` = 1, we get,
a2 = 16, b2 = 48
Slope of x + 2y – 7 = 0 is `-1/2`
The required tangent is perpendicular to it
∴ its slope = m = 2
∴ by (1), the required equations of tangents are
y = `2x ± sqrt(16(4) - 48)`
∴ y = 2x ± 4
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