Let m1 and m2 be the slopes of the tangents drawn from the point P(4,1) to the hyperbola H : \(\frac{y^2}{25} − \frac{x^2}{16 }\)= 1. If Q is the point from which the tangents drawn to H have slopes |m1| and |m2| and they make positive intercepts α and β on the x-axis, then \(\frac{( P Q ) ^2}{ α β}\) α β is equal to _____.
Correct Answer: 8
Solution and Explanation
Solution:
The equation of the tangent to the hyperbola \( \frac{y^2}{25} - \frac{x^2}{16} = 1 \) is given by:
\( y = mx \pm \sqrt{a^2 m^2 - b^2} \),
where \( a^2 = 25 \) and \( b^2 = 16 \).
For the point \( P(4, 1) \), substitute \( x = 4, y = 1 \):
\( 1 = 4m \pm \sqrt{25m^2 - 16} \).
Simplifying:
\( 1 - 4m = \pm \sqrt{25m^2 - 16} \).
Squaring both sides:
\( (1 - 4m)^2 = 25m^2 - 16 \).
Expanding:
\( 1 - 8m + 16m^2 = 25m^2 - 16 \).
Combining terms:
\( 9m^2 + 8m - 17 = 0 \).
Factoring:
\( (9m + 17)(m - 1) = 0 \).
Thus, \( m_1 = 1 \) and \( m_2 = -\frac{17}{9} \).
The equation of the tangent with \( m_1 = 1 \) is:
\( y = x - 3 \),
which intercepts the x-axis at \( \alpha = 3 \).
The equation of the tangent with \( m_2 = -\frac{17}{9} \) is:
\( y = -\frac{17}{9}x + \frac{77}{9} \),
which intercepts the x-axis at \( \beta = \frac{77}{17} \).
Intersection Point Q:
Solve the system of equations:
\( y = x - 3 \)
\( y = -\frac{17}{9}x + \frac{77}{9} \)
Equating the y-values:
\( x - 3 = -\frac{17}{9}x + \frac{77}{9} \)
\( 9x - 27 = -17x + 77 \)
\( 26x = 104 \)
\( x = 4 \)
Substitute \( x = 4 \) into \( y = x - 3 \):
\( y = 4 - 3 = 1 \)
So, \( Q = (4, 1) \).
Distance PQ:
\( PQ^2 = (4 - 4)^2 + (1 - 1)^2 = 0 \).
Calculate the ratio:
\( \frac{(PQ)^2}{\alpha \beta} = \frac{0}{3 \cdot \frac{77}{17}} = 0 \).
There seems to be an error in the original problem or the desired answer, as the final value is 0.
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