Q1) The equation of the normal to the curve: $$ x=6 t-5, y=t^{2} $$ at $$ t=1 $$ is ........ \begin{tabular}{ll|l|l|l|} (A) $$ \mid 3 y+x-4=0 $$ & B) $$ 3 x+y-4=0 $$ & (C) $$ \mid 3 y-x-2=0 $$ & (D) $$ \mid 3 x-y-2=0 $$ \ \hline\end{tabular}

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To find the equation of the normal to the curve defined by the parametric equations $$ x = 6t - 5 $$ and $$ y = t^2 $$ at $$ t = 1 $$, follow these steps:

1. **Determine the Point on the Curve at $$ t = 1 $$:**
   $$
   x = 6(1) - 5 = 1
   $$
   $$
   y = (1)^2 = 1
   $$
   So, the point is $$ (1, 1) $$.

2. **Find the Slope of the Tangent ($$ \frac{dy}{dx} $$):**
   $$
   \frac{dx}{dt} = 6
   $$
   $$
   \frac{dy}{dt} = 2t
   $$
   $$
   \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t}{6} = \frac{t}{3}
   $$
   At $$ t = 1 $$, the slope of the tangent is:
   $$
   \frac{dy}{dx} = \frac{1}{3}
   $$

3. **Determine the Slope of the Normal:**
   The normal is perpendicular to the tangent, so its slope is the negative reciprocal of the tangent's slope:
   $$
   \text{Slope of Normal} = -3
   $$

4. **Write the Equation of the Normal:**
   Using the point-slope form:
   $$
   y - y_1 = m(x - x_1)
   $$
   $$
   y - 1 = -3(x - 1)
   $$
   Simplifying:
   $$
   y - 1 = -3x + 3
   $$
   $$
   3x + y - 4 = 0
   $$

Thus, the equation of the normal is:

**Answer:** Option B) $$ 3 x + y -4=0 $$
The equation of the normal is $$ 3x + y - 4 = 0 $$.

Q1) The equation of the normal to the curve: at is …

(A) B) (C) (D)

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