Question

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 _____.

Answer 1

Correct Answer: 8

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.