Sigma Percentile
JEE Advanced 2010
LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Comprehension Passage

Tangents are drawn from the point to the ellipse touching the ellipse at points and .
Question 1:

The coordinates of and are

Select Answer:

Question 2:

The orthocenter of the triangle is

Select Answer:

Question 3:

The equation of the locus of the point whose distances from the point and the line are equal, is

Select Answer:

Visualized Solution

Visualizing the Setup

  • Ellipse:
  • External Point:
  • We need to find points of tangency and .

Chord of Contact

  • The line joining the points of tangency is the Chord of Contact.
  • Equation is given by .

Equation of

  • Substitute into :
  • Simplifying:

Finding Points and

  • Substitute into the ellipse equation.
  • Solving yields and .
  • Corresponding values: and .
  • Points are and .

Setting up

  • Vertices: , ,
  • We need to find the Orthocenter (intersection of altitudes).

Altitudes of

  • Side : Vertical line .
  • Altitude from to must be horizontal: .
  • Side : Slope is .
  • Altitude from to has slope .

Finding the Orthocenter

  • Altitude 1:
  • Altitude 2:
  • Intersection:
  • Orthocenter .

Defining the Locus

  • Locus of a point equidistant from and line .
  • Distance to Point = Distance to Line.
  • This is the exact definition of a Parabola.

Focus and Directrix

  • Focus: The point .
  • Directrix: The line with equation .
  • Let the moving point be .

Setting up the Equation

  • Distance to Focus squared:
  • Distance to Directrix squared:
  • Equating them:

Final Locus Equation

  • Expand:
  • Rearranging:

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The ellipse is defined by the equation:
Given the external point , we draw two tangents to the ellipse touching at points and . The line segment is the Chord of Contact.
Using the elegant theorem, we substitute into the ellipse equation to find the equation of the chord:
This linear equation serves as the backbone for our geometric analysis.

The Orthocenter

Finding Beauty in Verticality
We consider with vertices , , and .
Since and share the same -coordinate, the side is a vertical line. Consequently, the altitude dropped from vertex to must be a horizontal line.
Because this altitude passes through , its equation is simply:
Next, we find the altitude from to the line . Since the slope of () is , the slope of the perpendicular altitude is .
Using the point-slope form for :
Solving the system and , we find the orthocenter :
Thus, the orthocenter is .

The Locus

The Parabola's Secret
We seek the locus of a point equidistant from the focus and the directrix (). By definition, this locus is a parabola.
We set the squared distance to the focus equal to the squared distance to the directrix:
Expanding both sides, we obtain:
After careful algebraic simplification, we arrive at the final equation of the locus:

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