Question

The number of values of c such that the straight line \(y=4x+c\) touches the curve \(\frac{x^2}{4}+y^2=1\) is

• A. 0
• B. 1
• C. 2
• D. infinite

Hint:

Touch condition → discriminant zero.

Answer 1

The Correct Option is C

Step 1: Substitute \(y=4x+c\) in curve: \[ \frac{x^2}{4}+(4x+c)^2=1. \]
Step 2: This becomes quadratic in x: \[ \frac{x^2}{4}+16x^2+8cx+c^2-1=0. \] \[ \left(16+\frac14\right)x^2+8cx+(c^2-1)=0. \]
Step 3: For tangency, discriminant = 0: \[ \Delta=(8c)^2-4(64.25)(c^2-1)=0. \]
Step 4: Solve gives two real values of c. Hence number = 2 → (C).