Question

If the equation of the tangent of the hyperbola \( 5x^2 - 9y^2 - 20x - 18y - 34 = 0 \) which makes an angle \( 45^\circ \) with the positive X-axis in positive direction is \( x+by+c=0 \) then \( b^2+c^2 = \)

• A. 2 or 13
• B. 5 or 26
• C. 2 or 26
• D. 26 or 28

Hint:

1. Convert the hyperbola equation to standard form \( \frac{(x-h)^2}{A^2} - \frac{(y-k)^2}{B^2} = 1 \) by completing the square. Let \( X=x-h, Y=y-k \). 2. The slope of the tangent is \( m = \tan \alpha \). 3. Equation of tangent to \( \frac{X^2}{A^2} - \frac{Y^2}{B^2} = 1 \) with slope \(m\) is \( Y = mX \pm \sqrt{A^2m^2 - B^2} \). 4. Substitute back \(X, Y\) to get the equation in terms of \(x,y\). Compare with the given form \(x+by+c=0\) to find \(b,c\).

Answer 1

The Correct Option is C

First, convert the hyperbola equation to standard form.
\( 5x^2 - 20x - 9y^2 - 18y = 34 \) \( 5(x^2 - 4x) - 9(y^2 + 2y) = 34 \) \( 5(x^2 - 4x + 4) - 9(y^2 + 2y + 1) = 34 + 5(4) - 9(1) \) \( 5(x-2)^2 - 9(y+1)^2 = 34 + 20 - 9 = 45 \) Divide by 45: \[ \frac{(x-2)^2}{9} - \frac{(y+1)^2}{5} = 1 \] Let \( X = x-2 \) and \( Y = y+1 \).
The equation is \( \frac{X^2}{A^2} - \frac{Y^2}{B^2} = 1 \), with \( A^2=9 \implies A=3 \) and \( B^2=5 \implies B=\sqrt{5} \).
The tangent makes an angle of \( 45^\circ \) with the positive X-axis, so its slope is \( m = \tan 45^\circ = 1 \).
The equation of a tangent with slope \(m\) to the standard hyperbola \( \frac{X^2}{A^2} - \frac{Y^2}{B^2} = 1 \) is \( Y = mX \pm \sqrt{A^2m^2 - B^2} \).
Substitute \( m=1, A^2=9, B^2=5 \): \( Y = 1 \cdot X \pm \sqrt{9(1)^2 - 5} \) \( Y = X \pm \sqrt{9-5} = X \pm \sqrt{4} = X \pm 2 \).
So, the tangents in the (X,Y) system are \( Y = X+2 \) and \( Y = X-2 \).
This is \( X-Y+2=0 \) or \( X-Y-2=0 \).
Convert back to (x,y) system: \( X=x-2, Y=y+1 \).
Case 1: \( (x-2) - (y+1) + 2 = 0 \) \( x-2-y-1+2 = 0 \implies x-y-1=0 \).
Comparing with \( x+by+c=0 \): \( b=-1, c=-1 \).
Then \( b^2+c^2 = (-1)^2 + (-1)^2 = 1+1=2 \).
Case 2: \( (x-2) - (y+1) - 2 = 0 \) \( x-2-y-1-2 = 0 \implies x-y-5=0 \).
Comparing with \( x+by+c=0 \): \( b=-1, c=-5 \).
Then \( b^2+c^2 = (-1)^2 + (-5)^2 = 1+25=26 \).
The possible values for \( b^2+c^2 \) are 2 or 26.
This matches option (3).