This solution will guide you through finding a point on a parabola closest to a given point and then calculating its distance from the directrix of a different parabola. This problem combines concepts of analytical geometry (parabolas, directrix, distance formula) with differential calculus (minimization).

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1. Problem Analysis and Strategic Approach

The problem presents two distinct tasks:

1. Finding Point P: Determine the coordinates of a point $\mathrm{P}(h, k)$ on the parabola $x=4y^2$ that is nearest to the external point $\mathrm{Q}(0,33)$.
2. Calculating Distance from Directrix: Once $\mathrm{P}$ is found, calculate its perpendicular distance from the directrix of a second parabola, $y^2 = 4(x+y)$.

Our strategy will involve:

• Using calculus to minimize the distance (or its square) in Part 1.
• Converting the second parabola's equation to its standard form to identify its directrix in Part 2.
• Applying the distance formula between a point and a line.

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2. Part 1: Finding the Point P Nearest to Q

2.1 Key Concept: Distance Minimization To find the point on a curve nearest to an external point, we use the principle of minimization. The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula: $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ Minimizing $D$ is equivalent to minimizing $D^2$, which simplifies calculations by eliminating the square root. We achieve this by taking the derivative of $D^2$ with respect to a parameter and setting it to zero to find critical points.

2.2 Representing a General Point on the Parabola $x=4y^2$ Parametrically The first parabola is given by the equation $x = 4y^2$. To represent any point on this parabola, we can use a parameter. Let $y=t$. Then, substituting this into the parabola's equation, we get $x=4t^2$. So, any general point $\mathrm{P}$ on the parabola can be written in parametric form as $\mathrm{P}(4t^2, t)$. This choice of parameter simplifies differentiation later.

2.3 Setting up the Squared Distance Function $f(t)$ We want to find the point $\mathrm{P}(4t^2, t)$ that is nearest to $\mathrm{Q}(0, 33)$. Let $D$ be the distance between $\mathrm{P}$ and $\mathrm{Q}$. We will minimize $D^2$. $D^2 = (x_P - x_Q)^2 + (y_P - y_Q)^2$ Substitute the coordinates of $\mathrm{P}(4t^2, t)$ and $\mathrm{Q}(0,33)$: $D^2 = (4t^2 - 0)^2 + (t - 33)^2$ $D^2 = (4t^2)^2 + (t - 33)^2$ Expand the expression: $D^2 = 16t^4 + (t^2 - 66t + 1089)$ Let $f(t) = 16t^4 + t^2 - 66t + 1089$. Our goal is to find the value of $t$ that minimizes this function.

2.4 Minimizing $f(t)$ using Calculus To find the minimum value of $f(t)$, we differentiate $f(t)$ with respect to $t$ and set the derivative equal to zero. This identifies the critical points. $f'(t) = \frac{d}{dt}(16t^4 + t^2 - 66t + 1089)$ $f'(t) = 16(4t^3) + 2t - 66$ $f'(t) = 64t^3 + 2t - 66$ Set $f'(t) = 0$: $64t^3 + 2t - 66 = 0$ Divide the entire equation by 2 to simplify: $32t^3 + t - 33 = 0$ This is a cubic equation. For JEE problems, often there's a simple integer root that can be found by inspection. Let's test small integer values for $t$:

• If $t=0$, $32(0)^3 + 0 - 33 = -33 \neq 0$.
• If $t=1$, $32(1)^3 + 1 - 33 = 32 + 1 - 33 = 0$. So, $t=1$ is a root of the equation. This means $(t-1)$ is a factor of the cubic polynomial.

Tip for factoring cubics: Once an integer root is found, use polynomial division or synthetic division to find the other factors. Using synthetic division with $t=1$:

So, the cubic equation can be factored as: $(t-1)(32t^2 + 32t + 33) = 0$ Now we need to analyze the quadratic factor $32t^2 + 32t + 33$. We check its discriminant, $\Delta = b^2 - 4ac$, to see if it has any real roots. Here, $a=32$, $b=32$, $c=33$. $\Delta = (32)^2 - 4(32)(33)$ $\Delta = 1024 - 4224$ $\Delta = -3200$ Since the discriminant $\Delta < 0$, the quadratic equation $32t^2 + 32t + 33 = 0$ has no real roots. This implies that $t=1$ is the only real value of $t$ for which $f'(t)=0$.

To confirm that $t=1$ corresponds to a minimum, we can use the second derivative test. $f''(t) = \frac{d}{dt}(64t^3 + 2t - 66)$ $f''(t) = 192t^2 + 2$ Evaluate $f''(t)$ at $t=1$: $f''(1) = 192(1)^2 + 2 = 192 + 2 = 194$ Since $f''(1) = 194 > 0$, the function $f(t)$ has a local minimum at $t=1$. As it's the only real critical point, it represents the global minimum.

2.5 Determining the Coordinates of Point P Now that we have the value of $t$ that minimizes the distance, we substitute $t=1$ back into the parametric coordinates of $\mathrm{P}(4t^2, t)$: $P(4(1)^2, 1)$ $P(4, 1)$ So, the point $\mathrm{P}(h, k)$ on the parabola $x=4y^2$ nearest to $\mathrm{Q}(0,33)$ is $(4, 1)$. Thus, $h=4$ and $k=1$.

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3. Part 2: Distance of P from the Directrix of the Second Parabola

3.1 Key Concept: Standard Form of a Parabola and its Directrix To find the directrix of a parabola, we must first convert its equation into one of the standard forms:

• Horizontal Parabola: $(y-k)^2 = 4a(x-h)$
  • Vertex: $(h, k)$
  • Focal length: $|a|$
  • Opens right if $a>0$, left if $a<0$.
  • Directrix: $x = h - a$
• Vertical Parabola: $(x-h)^2 = 4a(y-k)$
  • Vertex: $(h, k)$
  • Focal length: $|a|$
  • Opens up if $a>0$, down if $a<0$.
  • Directrix: $y = k - a$

3.2 Converting the Equation $y^2 = 4(x+y)$ to Standard Form The equation of the second parabola is $y^2 = 4(x+y)$. Our goal is to rearrange this equation to match one of the standard forms, typically by completing the square for the variable that is squared. First, distribute the 4 on the right side: $y^2 = 4x + 4y$ Move all terms involving $y$ to one side and terms involving $x$ to the other side: $y^2 - 4y = 4x$ Now, complete the square for the terms involving $y$. To do this, take half of the coefficient of the $y$ term (which is $-4$), square it $(\frac{-4}{2})^2 = (-2)^2 = 4$, and add it to both sides of the equation: $y^2 - 4y + 4 = 4x + 4$ Factor the perfect square trinomial on the left side and factor out the common term on the right side: $(y-2)^2 = 4(x+1)$ This is now in the standard form $(y-k)^2 = 4a(x-h)$.

3.3 Identifying Parameters and the Directrix By comparing $(y-2)^2 = 4(x+1)$ with $(y-k)^2 = 4a(x-h)$:

• The vertex $(h, k)$ is $(-1, 2)$.
• The value of $4a$ is $4$, which implies $a=1$. Since the $y$ term is squared and $a=1$ (positive), this parabola opens to the right. For a parabola of the form $(y-k)^2 = 4a(x-h)$, the equation of the directrix is $x = h - a$. Substitute the values $h=-1$ and $