Question

If one end of the latus rectum of the parabola \(y^2=16x\) is \((4,8)\), then the coordinates of the other end of the latus rectum are

• A. \((4,-16)\)
• B. \((4,10)\)
• C. \((4,-10)\)
• D. \((4,16)\)
• E. \((4,-8)\)

Hint:

For the parabola \(y^2=4ax\), always remember that the endpoints of the latus rectum are \((a,2a)\) and \((a,-2a)\).

Answer 1

Answer: (E)

Step 1: Compare the given parabola with the standard form.
The standard form of a parabola is:
\[ y^2=4ax \] We are given:
\[ y^2=16x \] Comparing, we get:
\[ 4a=16 \Rightarrow a=4 \]

Step 2: Recall the latus rectum of the parabola \(y^2=4ax\).
For the parabola \(y^2=4ax\), the latus rectum is the line passing through the focus and perpendicular to the axis of the parabola. Its equation is:
\[ x=a \] So here, since \(a=4\), the latus rectum is:
\[ x=4 \]

Step 3: Recall the endpoints of the latus rectum.
For the parabola \(y^2=4ax\), the two endpoints of the latus rectum are:
\[ (a,2a) \quad \text{and} \quad (a,-2a) \] Since \(a=4\), these become:
\[ (4,8) \quad \text{and} \quad (4,-8) \]

Step 4: Use the point given in the question.
The question states that one end of the latus rectum is \((4,8)\). Therefore, the other end must be the symmetric point with respect to the \(x\)-axis.

Step 5: Find the symmetric point.
The symmetric point of \((4,8)\) about the \(x\)-axis is obtained by changing the sign of the \(y\)-coordinate while keeping the \(x\)-coordinate the same. Thus, we get:
\[ (4,-8) \]

Step 6: Verify that this point lies on the parabola.
Substitute \((4,-8)\) into the equation \(y^2=16x\):
\[ (-8)^2=16\cdot 4 \] \[ 64=64 \] So the point does lie on the parabola.

Step 7: State the final answer.
Hence, the coordinates of the other end of the latus rectum are:
\[ \boxed{(4,-8)} \] which matches option \((5)\).