Question

Let y = t² - 4t -10 and ax + by + c = 0 be the equation of the normal L. If G.C.D of (a,b,c) is 1, then m(a+b+c) =

• A. 8
• B. \(\frac{-64}{5}\)
• C. -8
• D. 5

Answer 1

The Correct Option is D

To solve the problem, we need to find the value of $m(a + b + c)$ for the normal line at the point $(1, 2)$ on the parametric curve defined by $x = t^2 - 7t + 7$ and $y = t^2 - 4t - 10$.

1. Find the Parameter $t$ at $(1, 2)$:
At $(1, 2)$, set $x = t^2 - 7t + 7 = 1$ and $y = t^2 - 4t - 10 = 2$.
Solve $t^2 - 7t + 6 = 0$:
$(t - 1)(t - 6) = 0$, so $t = 1$ or $t = 6$.
Solve $t^2 - 4t - 12 = 0$:
$(t - 6)(t + 2) = 0$, so $t = 6$ or $t = -2$.
The common solution is $t = 6$.

2. Compute the Derivative $\frac{dy}{dx}$:
Find $\frac{dx}{dt} = 2t - 7$ and $\frac{dy}{dt} = 2t - 4$.
At $t = 6$:
$\frac{dx}{dt} = 2(6) - 7 = 5$, $\frac{dy}{dt} = 2(6) - 4 = 8$.
Thus, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{8}{5}$.

3. Determine the Slope of the Normal:
The slope of the tangent is $\frac{8}{5}$, so the slope of the normal is $m = -\frac{5}{8}$.

4. Find the Equation of the Normal:
At $(1, 2)$, the normal’s equation is $y - 2 = -\frac{5}{8} (x - 1)$.
Multiply through by 8:
$8y - 16 = -5x + 5$.
Rearrange:
$5x + 8y - 21 = 0$.

5. Evaluate $m(a + b + c)$:
From $5x + 8y - 21 = 0$, we have $a = 5$, $b = 8$, $c = -21$.
The GCD of $(5, 8, -21)$ is 1, confirming the equation is in standard form.
Compute $m(a + b + c) = -\frac{5}{8} (5 + 8 - 21) = -\frac{5}{8} (-8) = 5$.

Final Answer:
The value of $m(a + b + c)$ is $5$.