<p>If the tangent and normal at a point on rectangular hyperbola cut-off intercept \(a_1, a_2\) on x-axis and \(b_1, b_2\) on the y-axis, then \(a_1a_2 + b_1b_2\) is equal to:</p>
Step-by-Step Solution
Key Concept: For a rectangular hyperbola xy = c², the tangent and normal at any point have slopes that are negative reciprocals. By finding the intercepts of both lines and using the relationship between their slopes, we can establish that the sum of products of intercepts equals zero.
**Step 1: Set up the rectangular hyperbola**
Consider a rectangular hyperbola given by $xy = c^2$.
Let $P(ct, c/t)$ be a point on the hyperbola.
**Step 2: Find the equation of the tangent at P**
Differentiating $xy = c^2$ with respect to $x$ yields $x \frac{dy}{dx} + y = 0$, so $\frac{dy}{dx} = -\frac{y}{x}$.
At the point $P(ct, c/t)$, the slope of the tangent is $m_T = -\frac{c/t}{ct} = -\frac{1}{t^2}$.
The equation of the tangent at $P$ is:
$$y - \frac{c}{t} = -\frac{1}{t^2}(x - ct)$$
Multiplying by $t^2$:
$$t^2y - ct = -x + ct$$
$$x + t^2y = 2ct$$
**Step 3: Find the intercepts of the tangent**
To find the x-intercept $a_1$, set $y=0$ in the tangent equation:
$$x + t^2(0) = 2ct \implies a_1 = 2ct$$
To find the y-intercept $b_1$, set $x=0$ in the tangent equation:
$$0 + t^2y = 2ct \implies b_1 = \frac{2ct}{t^2} = \frac{2c}{t}$$
**Step 4: Find the equation of the normal at P**
The slope of the normal is the negative reciprocal of the tangent's slope: $m_N = - \frac{1}{m_T} = - \frac{1}{-1/t^2} = t^2$.
The equation of the normal at $P$ is:
$$y - \frac{c}{t} = t^2(x - ct)$$
$$y = t^2x - ct^3 + \frac{c}{t}$$
Multiplying by $t$:
$$ty = t^3x - ct^4 + c$$
**Step 5: Find the intercepts of the normal**
To find the x-intercept $a_2$, set $y=0$ in the normal equation $ty = t^3x - ct^4 + c$:
$$0 = t^3x - ct^4 + c$$
$$t^3x = ct^4 - c$$
$$a_2 = \frac{c(t^4 - 1)}{t^3}$$
To find the y-intercept $b_2$, set $x=0$ in the normal equation $y = t^2x - ct^3 + \frac{c}{t}$:
$$b_2 = t^2(0) - ct^3 + \frac{c}{t}$$
$$b_2 = \frac{c}{t} - ct^3 = \frac{c(1 - t^4)}{t}$$
**Step 6: Calculate $a_1a_2 + b_1b_2$**
Substitute the expressions for $a_1, a_2, b_1, b_2$:
$$a_1a_2 = (2ct) \cdot \left(\frac{c(t^4 - 1)}{t^3}\right) = \frac{2c^2t(t^4 - 1)}{t^3} = \frac{2c^2(t^4 - 1)}{t^2}$$
$$b_1b_2 = \left(\frac{2c}{t}\right) \cdot \left(\frac{c(1 - t^4)}{t}\right) = \frac{2c^2(1 - t^4)}{t^2}$$
Now, sum these products:
$$a_1a_2 + b_1b_2 = \frac{2c^2(t^4 - 1)}{t^2} + \frac{2c^2(1 - t^4)}{t^2}$$
$$a_1a_2 + b_1b_2 = \frac{2c^2}{t^2} [(t^4 - 1) + (1 - t^4)]$$
$$a_1a_2 + b_1b_2 = \frac{2c^2}{t^2} [t^4 - 1 + 1 - t^4]$$
$$a_1a_2 + b_1b_2 = \frac{2c^2}{t^2} [0]$$
$$a_1a_2 + b_1b_2 = 0$$
Correct Answer: C