Let T be the tangent to the parabola y² = 16x at the point (64, 32). Let L be the tangent to the same parabola at another point (x₁, y₁) on the parabola. If L and T are perpendicular to each other, then the distance between the point (x₁, y₁) and the focus of the parabola, is

Correct Answer: Option 3: $\frac{17}{4}$

Let us solve this step-by-step.

Step 1: Understand the given parabola and its properties.
The equation of the parabola is:

$y^2=16x$

Comparing this with the standard equation of a parabola $y^2=4ax$, we find:
$4a=16\Rightarrow a=4$

The focus of this parabola is at the point $S(a,0)=(4,0)$.

Step 2: Find the slope of the tangent T.
Differentiating $y^2=16x$ with respect to $x$ gives:

$2y\frac{dy}{dx}=16\Rightarrow \frac{dy}{dx}=\frac{8}{y}$

At the point $(64,32)$, the slope of the tangent $T$, denoted as $m_T$, is:

$m_T=\frac{8}{32}=\frac{1}{4}$

Step 3: Find the slope of tangent L and the point (x₁, y₁).
Since tangent $L$ is perpendicular to tangent $T$, the product of their slopes must be $-1$:

$m_L·m_T=-1\Rightarrow m_L·\left(\frac{1}{4}\right)=-1\Rightarrow m_L=-4$

The slope of the tangent to the parabola at any point $(x_1,y_1)$ is given by $\frac{8}{y_1}$. Equating this to $m_L$:

$\frac{8}{y_1}=-4\Rightarrow y_1=-2$

Since $(x_1,y_1)$ lies on the parabola $y^2=16x$:

${(-2)}^2=16x_1\Rightarrow 4=16x_1\Rightarrow x_1=\frac{1}{4}$

So, the point $(x_1,y_1)$ is $\left(\frac{1}{4},-2\right)$.

Step 4: Calculate the distance from (x₁, y₁) to the focus.
For any point on the parabola $y^2=4ax$, the distance to the focus (focal distance) is given by the formula:

$\text{Focal Distance}=x+a$

Substituting $x_1=\frac{1}{4}$ and $a=4$:

$\text{Distance}=\frac{1}{4}+4=\frac{17}{4}$

Thus, the required distance is $\frac{17}{4}$.