Let a parabola $P$ be such that its vertex and focus lie on the positive x-axis at a distance 2 and 4 units from the origin, respectively. If tangents are drawn from $O(0, 0)$ to the parabola $P$ which meet $P$ at $S$ and $R$, then the area (in sq. units) of $\Delta SOR$ is equal to :
Show Hint
For a parabola \(y^2 = 4a(x-h)\), the tangents from the origin are symmetric if the vertex lies on the x-axis. This results in the triangle being isosceles with a vertical base, simplifying the area calculation.
Step 1: Understanding the Concept:
First, we find the standard equation of the parabola.
Then, we find the equations of the tangents from the origin and determine their points of contact.
Finally, the area of the triangle formed by the origin and the contact points is calculated. Step 2: Key Formula or Approach:
1. Parabola with vertex \((h, k)\): \((y - k)^2 = 4a(x - h)\).
2. Tangent condition: Discriminant of the combined line and curve equation must be zero.
3. Area of triangle with vertices \((0,0), (x_1, y_1), (x_2, y_2)\) is \(\frac{1}{2}|x_1 y_2 - x_2 y_1|\). Step 3: Detailed Explanation:
Vertex \(V = (2, 0)\), Focus \(F = (4, 0)\).
Distance \(a = 4 - 2 = 2\).
Parabola: \(y^2 = 4(2)(x - 2) \Rightarrow y^2 = 8x - 16\).
Let the tangent from origin be \(y = mx\).
Substitute: \((mx)^2 = 8x - 16 \Rightarrow m^2x^2 - 8x + 16 = 0\).
For tangency, \(D = 0 \Rightarrow (-8)^2 - 4(m^2)(16) = 0\).
\(64 - 64m^2 = 0 \Rightarrow m = \pm 1\).
Tangents are \(y = x\) and \(y = -x\).
Point of contact for \(y = x\): \(x^2 - 8x + 16 = 0 \Rightarrow (x-4)^2 = 0 \Rightarrow x = 4\).
So, \(S = (4, 4)\).
Point of contact for \(y = -x\): \((-x)^2 - 8x + 16 = 0 \Rightarrow (x-4)^2 = 0 \Rightarrow x = 4\).
So, \(R = (4, -4)\).
Area of \(\Delta SOR\) with \(O(0,0), S(4,4), R(4,-4)\):
Area $= \frac{1}{2} |4(-4) - 4(4)| = \frac{1}{2} |-16 - 16| = 16$. Step 4: Final Answer:
The area is 16 sq. units.
