The slope of the tangent to the hyperbola x225−y216=1 which passes through the point (1,4) is
A
−1
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B
−43
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C
2
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D
1
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Solution
The correct options are A1 B−43 The equation of a tangent is y=mx±√a2m2−b2 Hence, y=mx±√25m2−16 is the equation of a tangent to the hyperbola. This passes through (1,4), so 4=m±√25m2−16 ⇒16−8m+m2=25m2−16 ⇒24m2+8m−32=0 ⇒m=1,−43