Question

If the straight line \( y = 4x + c \) touches the ellipse \( \frac{x^2}{4} + y^2 = 1 \), then \( c \) is equal to:

• A. \( 0 \)
• B. \( \pm \sqrt{65} \)
• C. \( \pm \sqrt{62} \)
• D. \( \pm \sqrt{2} \)
• E. \( \pm 13 \)

Hint:

The condition \( c^2 = a^2 m^2 + b^2 \) is derived by substituting \( y = mx + c \) into the ellipse equation and setting the discriminant of the resulting quadratic equation in \( x \) to zero (since a tangent touches at exactly one point).

Answer 1

The Correct Option is B

Concept: A straight line \( y = mx + c \) is a tangent to (or "touches") an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) if and only if it satisfies the condition of tangency. This condition relates the slope of the line, the semi-axes of the ellipse, and the y-intercept of the line. The formula for the condition of tangency is: \[ c^2 = a^2 m^2 + b^2 \] Using this relationship, we can solve for the unknown constant \( c \).

Step 1: Identify the parameters from the given equations.
From the line \( y = 4x + c \):
• Slope \( m = 4 \) From the ellipse \( \frac{x^2}{4} + \frac{y^2}{1} = 1 \):
• \( a^2 = 4 \)
• \( b^2 = 1 \)

Step 2: Apply the condition of tangency.
Substitute the identified values into the formula \( c^2 = a^2 m^2 + b^2 \): \[ c^2 = (4)(4)^2 + 1 \] \[ c^2 = 4(16) + 1 \] \[ c^2 = 64 + 1 \] \[ c^2 = 65 \]

Step 3: Solve for \( c \).
Taking the square root of both sides: \[ c = \pm \sqrt{65} \] This means there are two possible lines with slope 4 that are tangent to the ellipse, one touching the upper half and one touching the lower half.