Question

The equation of the line tangent to the curve \(y^{2}=4x\) with an inclination of \(60^{\circ }\) is:

• A. \(3x-\sqrt{3}y-1=0\)
• B. \(3x-\sqrt{3}y+1=0\)
• C. \(3x+\sqrt{3}y+1=0\)
• D. \(3x+\sqrt{3}y-1=0\)

Hint:

Use the tangent y = mx + 1/m for y^2 = 4x with m = tan 60 degrees.

Answer 1

The Correct Option is B

Step 1: Substitute into the parabola
Let the line be $y=\sqrt{3}x+c$. Putting in $y^{2}=4x$: $3x^{2}+(2\sqrt{3}c-4)x+c^{2}=0$.

Step 2: Tangency condition
The discriminant must be zero: $(2\sqrt{3}c-4)^{2}=12c^{2}\Rightarrow 16-16\sqrt{3}c=0\Rightarrow c=\frac{1}{\sqrt{3}}$.

Step 3: Result
The line is $y=\sqrt{3}x+\frac{1}{\sqrt{3}}$, i.e. $3x-\sqrt{3}y+1=0$.

Final Answer:
The line is 3x - sqrt3 y + 1 = 0. \[ \boxed{\text{(B) }3x-\sqrt{3}y+1=0} \]