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If the tangent and normal to the hyperbo...

If the tangent and normal to the hyperbola `x^(2) - y^(2) = 4` at a point cut off intercepts `a_(1)` and `a^(2)` respectively on the x-axis, and `b_(1)` and `b_(2)` respectively on the y-axis, then the value of `a_(1)a_(2) + b_(1)b_(2)` is

A

`-1`

B

0

C

4

D

1

Text Solution

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The correct Answer is:
To solve the problem, we need to find the values of \( a_1, a_2, b_1, \) and \( b_2 \) based on the tangent and normal to the hyperbola \( x^2 - y^2 = 4 \) at a point \( P \) on the hyperbola. ### Step 1: Identify the point on the hyperbola Let the point \( P \) on the hyperbola be represented in terms of a parameter \( \theta \): \[ P = (2 \sec \theta, 2 \tan \theta) \] Here, we have used the standard form of the hyperbola \( \frac{x^2}{4} - \frac{y^2}{4} = 1 \). **Hint:** Use the parametric equations of the hyperbola to express the coordinates of point \( P \). ### Step 2: Find the equation of the tangent The equation of the tangent to the hyperbola at point \( P \) is given by: \[ x \sec \theta - y \tan \theta = 2 \] **Hint:** Recall the formula for the tangent to a hyperbola at a given point. ### Step 3: Find the x-intercepts \( a_1 \) and \( a_2 \) To find the x-intercept \( a_1 \), set \( y = 0 \) in the tangent equation: \[ x \sec \theta = 2 \implies a_1 = 2 \cos \theta \] To find the x-intercept \( a_2 \) of the normal, we first write the equation of the normal: \[ \frac{4}{2} \cdot \frac{x}{2 \sec \theta} + \frac{4}{2} \cdot \frac{y}{2 \tan \theta} = 4 \] This simplifies to: \[ x \sec \theta + y \tan \theta = 4 \] Setting \( y = 0 \): \[ x \sec \theta = 4 \implies a_2 = 4 \cos \theta \] **Hint:** To find intercepts, substitute \( y = 0 \) in the tangent and normal equations. ### Step 4: Find the y-intercepts \( b_1 \) and \( b_2 \) For the y-intercept \( b_1 \) of the tangent, set \( x = 0 \): \[ -y \tan \theta = 2 \implies b_1 = -\frac{2}{\tan \theta} = -2 \cot \theta \] For the y-intercept \( b_2 \) of the normal, set \( x = 0 \): \[ y \tan \theta = 4 \implies b_2 = 4 \tan \theta \] **Hint:** Similarly, set \( x = 0 \) in the tangent and normal equations to find y-intercepts. ### Step 5: Compute \( a_1 a_2 + b_1 b_2 \) Now we can compute: \[ a_1 a_2 = (2 \cos \theta)(4 \cos \theta) = 8 \cos^2 \theta \] \[ b_1 b_2 = (-2 \cot \theta)(4 \tan \theta) = -8 \] Thus, \[ a_1 a_2 + b_1 b_2 = 8 \cos^2 \theta - 8 \] **Hint:** Use trigonometric identities to simplify the expression. ### Step 6: Simplify the expression Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \): \[ 8 \cos^2 \theta - 8 = 8(1 - \sin^2 \theta) - 8 = 0 \] ### Final Answer The value of \( a_1 a_2 + b_1 b_2 \) is: \[ \boxed{0} \]
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