Question

A normal is drawn at the point \( P \) to the parabola \( y^2 = 8x \), which is inclined at \( 60^\circ \) with the straight line \( y = 8 \). Then the point \( P \) lies on the straight line:

• A. \( 2x + y - 12 - 4\sqrt{3} = 0 \)
• B. \( 2x - y - 12 + 4\sqrt{3} = 0 \)
• C. \( 2x - y - 12 - 4\sqrt{3} = 0 \)
• D. None of these

Hint:

In problems involving normals to curves, first find the slope of the tangent at the given point, then use the fact that the slope of the normal is the negative reciprocal. Use the angle information to find the equation of the normal.

Answer 1

The Correct Option is C

The equation of the given parabola is: \[ y^2 = 8x \] The slope of the tangent to the parabola at any point \( (x_1, y_1) \) on the parabola is found by differentiating the equation implicitly: \[ 2y \frac{dy}{dx} = 8 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{4}{y} \] So, the slope of the tangent at \( (x_1, y_1) \) is \( \frac{4}{y_1} \). The slope of the normal at this point is the negative reciprocal of the tangent slope: \[ {Slope of the normal} = -\frac{y_1}{4} \] Now, we are given that the normal is inclined at \( 60^\circ \) with the horizontal line \( y = 8 \), which has a slope of 0. The angle between the normal and the horizontal line is given by: \[ \tan(60^\circ) = \sqrt{3} \] Thus, the slope of the normal can also be expressed as \( \sqrt{3} \). Therefore: \[ -\frac{y_1}{4} = \sqrt{3} \] Solving for \( y_1 \): \[ y_1 = -4\sqrt{3} \] Substitute this value of \( y_1 \) into the equation of the parabola: \[ y_1^2 = 8x_1 \quad \Rightarrow \quad (-4\sqrt{3})^2 = 8x_1 \] \[ 48 = 8x_1 \quad \Rightarrow \quad x_1 = 6 \] So, the point \( P \) is \( (6, -4\sqrt{3}) \). Now, to find the straight line that passes through this point, we substitute \( x = 6 \) and \( y = -4\sqrt{3} \) into the equation of the straight line. The line that passes through this point is: \[ 2x - y - 12 - 4\sqrt{3} = 0 \] Thus, the correct answer is Option C.