1998

JEE Main 1998 Mathematics question on Conic Sections

The number of values of cc such that the straight line y=4x+cy = 4x + c touches the curve x24+y2=1\frac{x^2}{4} + y^2 = 1 is
  1. A.
    0
  2. B.
    2
  3. C.
    1
  4. D.
    ∞\infty
Show answer and solution

Correct option: B

Solution

For the line y = 4x + c to touch the ellipse x²/4 + y²= 1, substitute y = 4x + c into the ellipse equation and require the discriminant to be zero.

Substituting: x²/4 + (4x+c)² = 1 → x²/4 + 16x² + 8cx + c² = 1 → (65/4)x² + 8cx + (c²-1) = 0.

Setting discriminant = 0: (8c)² - 4·(65/4)·(c²-1) = 0 → 64c² - 65(c²-1) = 0 → -c² + 65 = 0 → c² = 65, giving exactly c = ±√65.

These are 2 distinct values of c.

[1998]

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