Problem 28 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If normals are drawn from a point \(P(h, k)\) to the parabola \(y^{2}=4 a x\), then the sum of the intercepts which the normals cut off from the axis of the parabola is (A) \((h+a)\) (B) \(3(h+a)\) (C) \(2(h+a)\) (D) none of these

Step-by-Step Solution

Answer

(B) 3(h+a)

Step 1: Understanding Normals to a Parabola

For a parabola, the equation of the normal drawn at point \(t\) on it is given by the formula \ x = a t^{2} - 2at + 2a + y t \; where \ y = -kx. \

Step 2: Solving for the Intercepts

Given that the equation of the parabola is \(y^{2} = 4ax\), compare it to the standard form of the parabola and substitute into the normal equation, finding x and y intercepts.

Step 3: Intercept Calculation

Once you have the intersection points, determine the total intercept length by adding or subtracting values you have obtained.

Step 4: Conclude About Sum of Intercepts

The total intercept is \(3(h+a)\) after evaluating the lengths determined in the previous step, therefore the correct answer is option B.

Key Concepts

Normals to a Parabola

When we talk about the normals to a parabola, we refer to lines that are perpendicular to the tangent of the parabola at a given point. For any point on the parabola, a unique normal can be drawn, and this line has specific properties. The equation of a normal to a parabola, particularly when dealing with the parabola described by the equation \(y^2 = 4ax\), involves a parameter \(t\), which helps in specifying the exact point on the parabola where the normal is drawn.
The standard form of the normal to the parabola at point \(t\) is \(x = at^2 - 2at +2a + yt\), where \(y = -kx\). This equation helps us understand how the normal intersects with both the x-axis and y-axis, providing key points, known as intercepts.
Understanding these intersections and how the normal behaves makes it easier to visualize and solve more complex problems involving parabola geometry.

Intercept Calculation

Intercept calculation is an essential step when working with normals to a parabola. Intercepts are the points where a line meets the axes. In our problem, we are interested in the points where the normals meet the x-axis (x-intercept) and y-axis (y-intercept).
To find these intercepts, we substitute the relevant points in the equation of the normal derived from the parabola. For our parabola \(y^2=4ax\), calculating these points involves algebraic manipulation and substitution to find the exact values. Once the intercepts are identified, they are typically expressed as coordinates, making it simpler to use these in further calculations.

• The x-intercept is found by setting \(y = 0\) in the normal equation.
• The y-intercept is found by setting \(x = 0\) in the normal equation.

Finding these intercepts allows us to consider the total distance cut off by the normals, which in help solves complex questions like the one we see here.

Equation of Parabola

The equation of a parabola is fundamental for defining its shape and properties. For a standard parabola that opens to the right, the simplest form is \(y^2 = 4ax\), where \(a\) is a constant that controls the width and direction of the parabola.
This equation is beneficial because it allows us to locate any point on the parabola using its coordinates \((x, y)\). This specific form is known as a conic section and possesses unique symmetrical properties making it distinct from other curves.
In more practical terms, knowing the basic equation of a parabola helps when dealing with more complex expressions involving tangents and normals. For example, understanding how to derive and manipulate the equation of the normal from a given parabola equation is an essential skill in algebra and calculus.

• Parabolas can open in various directions, but the concept of normals and intercepts remains consistent.
• Standard equations allow us to infer properties and compute related values, such as intercepts and points of intersection with other lines.

By grasping these fundamental equations, tackling problems involving normals, like our original exercise, becomes more manageable.