Sigma Percentile
JEE Main 2021 (February) (25 Feb Shift 2)
LEVELJEE Advanced

Animated Solution for Mathematics - Differentiation: If the curves and cut at right angles, then is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Curves

  • Let the given curves be and .
  • They intersect at a point .

Condition for Orthogonality

  • The curves cut at right angles (orthogonally).
  • This means their tangents at point are perpendicular.
  • Condition:

Slope of First Curve ()

  • Curve 1:
  • Differentiating with respect to :

Slope of Second Curve ()

  • Curve 2:
  • Differentiating with respect to :

Applying Orthogonality Condition

  • Substitute and into :

Simplifying the Equation

  • Cancel the negative signs on both sides.
  • Cancel one from numerator and denominator:

Solving for

  • We have and we know .
  • Substitute :

Analyzing the Target Expression

  • We need to find the value of .
  • From the second curve, we know .
  • Therefore,

Expressing Target in terms of

  • Target:
  • Substitute :
  • Target becomes:

Final Calculation

  • We have:
  • Rewrite using :
  • Substitute :
  • Final Answer: 4

The Sigma Insight: Tangents, Normals and Rate Measure

Solution Diagram

Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the elegant dance between two curves. When we look at the equations and , we aren't just looking at algebraic expressions.
We are looking at a sideways parabola and a rectangular hyperbola, two distinct geometric entities meeting at a point . The problem asks us to consider the case where these curves cut at right angles—a condition known as orthogonality.
This is a beautiful concept. It means that at the very moment they touch, their paths are perpendicular. Let us unravel this step by step.

The Calculus of Slopes

To understand the intersection, we must understand the direction of each curve at point . We need the slopes of the tangents.
For our first curve, , we differentiate with respect to :
Since the slope is defined as , we take the reciprocal:
Now, consider the second curve, . To find its slope , we use implicit differentiation with respect to :
We now have the "DNA" of both curves at the point of intersection.

The Orthogonality Condition

If two curves are orthogonal, their tangents are perpendicular. In the coordinate plane, this implies that the product of their slopes must be .
Setting , we substitute our expressions:
The negative signs cancel out, and the terms simplify to in the denominator. We are left with the following constraint:

The Algebraic Finale

We need to evaluate the expression . Since , our target is , which expands to .
We know from our first curve that . Substituting this into our constraint :
Now, we express in terms of . Since , then .
Substituting this into our target expression :
We rewrite as . Substituting into the expression:
The final result is 4.

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