Question

If the tangent drawn at \( A(2, 1) \) to the curve \( x = 1 + \frac{1}{y^2} \) meets the curve again at \( B \), then

• A. the tangent drawn at \( B \) coincides with the tangent drawn at \( A \)
• B. the angle between the tangents drawn at \( A \) and \( B \) is neither \( 0 \) nor \( \frac{\pi}{2} \)
• C. the tangent drawn at \( A \) and the tangent drawn at \( B \) are perpendicular to each other
• D. the tangent drawn at \( A \) is parallel to the tangent drawn at \( B \)

Hint:

Find the equation of the tangent at the given point \( A \). Find the intersection point \( B \) of this tangent with the curve. Calculate the slopes of the tangents at \( A \) and \( B \). Use the formula for the angle between two lines to determine the relationship between the tangents.

Answer 1

The Correct Option is B

The equation of the curve is \( x = 1 + \frac{1}{y^2} \).
Differentiating with respect to \( y \): $$ \frac{dx}{dy} = -\frac{2}{y^3} $$ So, \( \frac{dy}{dx} = -\frac{y^3}{2} \).
At point \( A(2, 1) \), the slope of the tangent is \( m_A = \frac{dy}{dx}\Big|_{(2, 1)} = -\frac{1^3}{2} = -\frac{1}{2} \).
The equation of the tangent at \( A(2, 1) \) is \( y - 1 = -\frac{1}{2}(x - 2) \) \( 2y - 2 = -x + 2 \) \( x + 2y - 4 = 0 \) To find the point \( B \) where the tangent meets the curve again, we substitute \( x = 4 - 2y \) into the curve equation \( x = 1 + \frac{1}{y^2} \): \( 4 - 2y = 1 + \frac{1}{y^2} \) \( 3 - 2y = \frac{1}{y^2} \) \( 3y^2 - 2y^3 = 1 \) \( 2y^3 - 3y^2 + 1 = 0 \) We know that \( y = 1 \) is a root (corresponding to point \( A \)).
So \( (y - 1) \) is a factor.
\( (y - 1)(2y^2 - y - 1) = 0 \) \( (y - 1)(2y + 1)(y - 1) = 0 \) The roots are \( y = 1 \) (twice) and \( y = -\frac{1}{2} \).
For point \( B \), \( y = -\frac{1}{2} \).
The corresponding \( x \) coordinate of \( B \) is \( x = 1 + \frac{1}{(-\frac{1}{2})^2} = 1 + \frac{1}{\frac{1}{4}} = 1 + 4 = 5 \).
So, \( B = (5, -\frac{1}{2}) \).
The slope of the tangent at \( B(5, -\frac{1}{2}) \) is \( m_B = \frac{dy}{dx}\Big|_{(5, -\frac{1}{2})} = -\frac{(-\frac{1}{2})^3}{2} = -\frac{-\frac{1}{8}}{2} = \frac{1}{16} \).
The angle \( \theta \) between the tangents at \( A \) and \( B \) is given by \( \tan \theta = \left| \frac{m_B - m_A}{1 + m_A m_B} \right| \).
\( \tan \theta = \left| \frac{\frac{1}{16} - (-\frac{1}{2})}{1 + (-\frac{1}{2})(\frac{1}{16})} \right| = \left| \frac{\frac{1}{16} + \frac{8}{16}}{1 - \frac{1}{32}} \right| = \left| \frac{\frac{9}{16}}{\frac{31}{32}} \right| = \left| \frac{9}{16} \cdot \frac{32}{31} \right| = \left| \frac{18}{31} \right| = \frac{18}{31} \).
Since \( \tan \theta = \frac{18}{31} \neq 0 \) and \( \tan \theta \) is finite and non-zero, the angle \( \theta \) is neither \( 0 \) nor \( \frac{\pi}{2} \).