A double orinate of the parabola y^2=2ax is of length 4a prove that the line joíning the vertex to its ends are right angle with figure

To tackle the problem of proving that the lines joining the vertex of the parabola \( y^2 = 2ax \) to the ends of a double ordinate of length \( 4a \) are at right angles, we first need to understand the geometry of the parabola and the concept of a double ordinate.

Understanding the Parabola

The equation \( y^2 = 2ax \) describes a parabola that opens to the right. The vertex of this parabola is at the origin, \( (0, 0) \). The focus of the parabola is located at \( (a, 0) \), and the directrix is the line \( x = -a \).

Defining a Double Ordinate

A double ordinate is a line segment that is perpendicular to the axis of symmetry of the parabola and intersects it at two points. For our parabola, the axis of symmetry is the x-axis. The double ordinate will have endpoints at points \( (x_1, y_1) \) and \( (x_1, -y_1) \) for some value of \( x_1 \).

Finding the Length of the Double Ordinate

The length of the double ordinate can be calculated using the formula for the y-coordinates derived from the parabola's equation:

• From \( y^2 = 2ax_1 \), we find \( y = \sqrt{2ax_1} \) and \( y = -\sqrt{2ax_1} \).
• The length of the double ordinate is given by \( |y_1 - (-y_1)| = 2|y_1| = 2\sqrt{2ax_1} \).

According to the problem, this length is equal to \( 4a \). Therefore, we set up the equation:

\( 2\sqrt{2ax_1} = 4a \)

Dividing both sides by 2 gives:

\( \sqrt{2ax_1} = 2a \)

Squaring both sides results in:

\( 2ax_1 = 4a^2 \)

From this, we can solve for \( x_1 \):

\( x_1 = \frac{4a^2}{2a} = 2a \)

Coordinates of the Ends of the Double Ordinate

Now that we have \( x_1 = 2a \), we can find the coordinates of the endpoints of the double ordinate:

• At \( x = 2a \), the y-coordinates are \( y = \sqrt{2a(2a)} = \sqrt{4a^2} = 2a \) and \( y = -\sqrt{4a^2} = -2a \).

Thus, the endpoints of the double ordinate are \( (2a, 2a) \) and \( (2a, -2a) \).

Proving the Right Angle

Next, we need to show that the lines joining the vertex \( (0, 0) \) to the endpoints \( (2a, 2a) \) and \( (2a, -2a) \) are perpendicular. The slopes of these lines can be calculated as follows:

• The slope of the line from \( (0, 0) \) to \( (2a, 2a) \) is \( \frac{2a - 0}{2a - 0} = 1 \).
• The slope of the line from \( (0, 0) \) to \( (2a, -2a) \) is \( \frac{-2a - 0}{2a - 0} = -1 \).

Two lines are perpendicular if the product of their slopes is \( -1 \). Here, we have:

\( 1 \times (-1) = -1 \)

This confirms that the lines are indeed perpendicular.

Visual Representation

To visualize this, imagine the parabola opening to the right with the vertex at the origin. The double ordinate stretches vertically at \( x = 2a \), creating a symmetrical shape. The lines drawn from the vertex to the endpoints of the double ordinate form a right angle, confirming our proof.

In summary, we have shown that the lines joining the vertex of the parabola to the ends of a double ordinate of length \( 4a \) are at right angles by calculating the slopes of these lines and demonstrating their product equals \( -1 \).