Problem 119 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(y=m x+\mathrm{c}\) is the normal at a point on the parabola \(\mathrm{y}^{2}=8 x\) whose focal distance is 8 units, then \(|\mathrm{c}|\) is equal to: (a) \(2 \sqrt{3}\) (b) \(8 \sqrt{3}\) (c) \(10 \sqrt{3}\) (d) \(16 \sqrt{3}\)

Step-by-Step Solution

Answer

\(|c| = 10\sqrt{3}\). The correct answer is (c).

Step 1: Understand the problem

We need to find the value of \(|c|\) when the line \(y = mx + c\) is the normal to the parabola \(y^2 = 8x\) at a certain point and the focal distance of this point is given as 8 units.

Step 2: Determine the focus of the parabola

For the parabola \(y^2 = 8x\), the form \(y^2 = 4ax\) gives us \(4a = 8\), thus \(a = 2\). The focus of this parabola is at \((a, 0) = (2, 0)\).

Step 3: Calculate the parametric point on the parabola

The general parametric point on the parabola, for parameter \(t\), is \((a t^2, 2at)\). Thus, substituting \(a = 2\), it becomes \((2t^2, 4t)\).

Step 4: Compute the focal distance for the parametric point

The focal distance is the distance from the point \((2t^2, 4t)\) to the focus \((2, 0)\). This distance is calculated using \ \[\sqrt{(2t^2 - 2)^2 + (4t - 0)^2} = \sqrt{(2t^2 - 2)^2 + 16t^2}.\] We set this distance equal to 8: \ \[\sqrt{(2t^2 - 2)^2 + 16t^2} = 8.\]

Step 5: Solve the equation for the parametric value

Square both sides to remove the square root:\[(2t^2 - 2)^2 + 16t^2 = 64.\] Expand: \[(4t^4 - 8t^2 + 4) + 16t^2 = 64.\] Simplify to obtain \[4t^4 + 8t^2 + 4 = 64.\] Thus, \[4t^4 + 8t^2 - 60 = 0.\] Divide by 4: \[t^4 + 2t^2 - 15 = 0.\] Solving for \(t^2\), we have \(t^2 = 3\). Thus, \(t = \pm \sqrt{3}\).

Step 6: Find the point of normal intersection

Using \(t = \sqrt{3}\), the point on the parabola is \((6, 4\sqrt{3})\).

Step 7: Derive the slope of the normal

For the parabola \(y^2 = 8x\), the slope of the tangent at any point \((2t^2, 4t)\) is \(\frac{a}{t}\), thus the slope of the normal is \(-t\). Substituting \(t = \sqrt{3}\), the slope of the normal is \(-\sqrt{3}\).

Step 8: Establish the normal equation

The equation of the normal is \(y - 4\sqrt{3} = -\sqrt{3}(x - 6)\). Expanding and rearranging, you find: \ \[y = -\sqrt{3} x + 6\sqrt{3} + 4\sqrt{3}.\] Simplify to get \[y = -\sqrt{3} x + 10\sqrt{3}.\] So \(c = 10\sqrt{3}\).

Step 9: Determine the absolute value of c

The problem asks for \(|c|\), which in this case is \(10\sqrt{3}\) since \(10\sqrt{3} > 0\).

Key Concepts

Focal Distance

In the context of parabolas, the focal distance is a critical concept as it describes the distance from any point on the parabola to its focus. For the equation \(y^2 = 8x\), we determine the focus by rewriting it in the standard form \(y^2 = 4ax\), leading to \(a = 2\). The focus therefore is located at \((2,0)\).

When calculating focal distance, we utilize the distance formula. For any given point \((x_1, y_1)\), its distance from the focus \((a,0)\) is given by:
\[\text{Distance} = \sqrt{(x_1 - a)^2 + (y_1 - 0)^2} \]

In our example, the focal distance is set to be 8 units, imposing a specific constraint on our calculation. Solving for this constraint helps us find particular characteristics of the parabola, such as its points or properties related to normals and tangents.

Normal to Parabola

The normal to a parabola is a line that is perpendicular to the tangent at any point on the parabola. Understanding this helps us analyze geometric features like direction and position of such lines in relation to the parabola.

For a parabola described by the equation \(y^2 = 8x\), the slope of the normal at any point \((2t^2, 4t)\) can be calculated by finding the negative reciprocal of the slope of the tangent. The slope of the tangent, given \( \frac{d}{dx}(y^2 = 8x)\), is affected by the parameter \(t\).

• Tangent slope: \(\frac{a}{t}\)
• Normal slope: \(-t\)

Thus, the normal line equation is derived by substituting these values into the general line form, resulting in normal lines that showcase the perpendicular relation to the tangent line.

For the calculated point \((6, 4\sqrt{3})\), the equation of the normal line becomes \(y - 4\sqrt{3} = -\sqrt{3}(x - 6)\). After simplifying, this translates to \(y = -\sqrt{3}x + 10\sqrt{3}\).

Parametric Equation

The parametric equations for a parabola provide an alternative way to express points on the curve using a parameter \(t\), rather than x and y alone. This approach simplifies many calculations, especially those involving geometry such as tangents and normals.

For the parabola \(y^2 = 8x\), we rearrange it to the standard form \(y^2 = 4ax\) to determine \(a = 2\). Utilizing the parameter \(t\), the parametric equations become:

• \(x = 2t^2\)
• \(y = 4t\)

This representation gives us a straightforward way to describe any point on the parabola, simplifying the calculations for focal distance, normals, and tangents.

By setting specific values for \(t\) such as \(t = \sqrt{3}\), we can directly find points like \((6, 4\sqrt{3})\) on the parabola, which are pivotal for solving related equations involving distances or angles.

Slope

Slope is the measure of steepness or incline of a line, crucial when dealing with tangents and normals. In parabolic contexts, slope helps determine how these lines interact with the curve.

For the parabola \(y^2 = 8x\), we calculate the slope of the tangent line at a point \((2t^2, 4t)\) using the derivative of the parabola’s equation. The slope of the tangent is \(\frac{1}{2t}\).

• The tangent slope provides insight into how sharply the line insinuates against the curve.
• The normal slope is determined by taking the negative reciprocal of the tangent slope, given as \(-t\).

This negative reciprocal relationship is foundational in identifying the normal line as perpendicular to the tangent. In practical steps, solving such slopes ensures we can correctly calculate line equations and further properties related to the parabola.

These slope calculations connect deeply with the geometric and algebraic understanding required to solve more complex parabolic problems, such as identifying specific line intersections or angles formed.