JEE Advanced level
A JEE Advanced-level example on Conic Sections (Parabola, Ellipse, Hyperbola)
Question. A tangent to the parabola y² = 8x makes an angle of 45° with the positive x-axis. Find the equation of this tangent, the point of contact on the parabola, and the area of the triangle formed by this tangent, the x-axis, and the latus rectum of the parabola.
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- For y² = 8x, we get 4a = 8, so a = 2. The tangent with slope m to y² = 4ax is y = mx + a/m. With m = tan(45°) = 1, the tangent is y = x + 2.
- Point of contact for tangent y = mx + a/m is (a/m², 2a/m). Substituting a = 2 and m = 1 gives point of contact as (2, 4).
- Tangent y = x + 2 meets the x-axis (y = 0) at point A = (−2, 0).
- The latus rectum is the line x = 2. It meets the x-axis at B = (2, 0) and meets the tangent at C = (2, 4).
- Triangle ABC has vertices (−2, 0), (2, 0), and (2, 4). Base AB = 4 units along x-axis, height = 4 units. Area = (1/2) × 4 × 4 = 8 square units.
Answer: Tangent: y = x + 2, Point of contact: (2, 4), Area of triangle: 8 square units
This example was solved twice independently and both solutions agree.