A tangent and a normal are drawn at the point $P(2, -4)$ on the parabola $y^2 = 8x$, which meet the directrix of the parabola at the points A and B respectively. If $Q(a, b)$ is a point such that $AQBP$ is a square, then $2a + b$ is equal to :
Show Hint
Remember that the diagonals of a square bisect each other. If the coordinates of three vertices are known and the shape is a square, the fourth vertex can be found quickly by using the midpoint property of diagonals.
Step 1: Understanding the Concept:
This question combines properties of parabolas (tangents, normals, and directrix) with the properties of a square. A square's diagonals are equal, bisect each other, and are perpendicular. Step 2: Key Formula or Approach:
For $y^2 = 4ax$, tangent at $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.
Directrix of $y^2 = 4ax$ is $x = -a$. Step 3: Detailed Explanation:
Given parabola is $y^2 = 8x$, so $4a = 8 \implies a = 2$.
Directrix is $x = -2$.
Point $P$ is $(2, -4)$. Tangent at P:
\[ y(-4) = 4(x + 2) \implies -4y = 4x + 8 \implies x + y + 2 = 0 \]
Intersection with directrix $x = -2$:
\[ -2 + y + 2 = 0 \implies y = 0 \]
So, point $A = (-2, 0)$. Normal at P:
The slope of the tangent is $-1$, so the slope of the normal is $1$.
\[ y - (-4) = 1(x - 2) \implies y + 4 = x - 2 \implies x - y - 6 = 0 \]
Intersection with directrix $x = -2$:
\[ -2 - y - 6 = 0 \implies y = -8 \]
So, point $B = (-2, -8)$. AQBP is a square:
In a square, the midpoints of the diagonals coincide. Diagonals are AB and PQ.
Midpoint of AB $= (\frac{-2-2}{2}, \frac{0-8}{2}) = (-2, -4)$.
Let $Q = (a, b)$. Midpoint of PQ must be $(-2, -4)$:
\[ \frac{a + 2}{2} = -2 \implies a + 2 = -4 \implies a = -6 \]
\[ \frac{b - 4}{2} = -4 \implies b - 4 = -8 \implies b = -4 \]
So, $Q = (-6, -4)$.
Now, $2a + b = 2(-6) + (-4) = -12 - 4 = -16$. Step 4: Final Answer:
The value of $2a + b$ is $-16$.
