Question

A normal chord PQ drawn at a point P on the parabola \(y^2=5x\) subtends a right angle at the vertex. If P lies in the first quadrant, then the other end Q of the normal chord is

• A. \( (\frac{5}{4}, \frac{5}{2}) \)
• B. (5, -5)
• C. \( (10, -5\sqrt{2}) \)
• D. \( (\frac{5}{2}, \frac{5\sqrt{2}}{2}) \)

Hint:

For a parabola \(y^2=4ax\), memorize these key results for a normal at point 't': 1. It meets the parabola again at \(t' = -t - 2/t\). 2. The normal chord subtends a right angle at the vertex if \(t^2=2\). 3. The normal passes through the focus if \(t^2=1\).

Answer 1

The Correct Option is C

For the parabola \(y^2 = 4ax\), the condition that the normal chord at a point \(t\) subtends a right angle at the vertex is \(t^2 = 2\).
Here, the parabola is \(y^2 = 5x\), so \(4a = 5 \implies a = 5/4\).
The point P is parameterized by \(t\). Since P is in the first quadrant, its y-coordinate, \(2at\), must be positive. This means \(t\) must be positive.
From \(t^2=2\), we choose \(t = \sqrt{2}\).
The coordinates of P are \( (at^2, 2at) = (\frac{5}{4}(2), 2(\frac{5}{4})\sqrt{2}) = (\frac{5}{2}, \frac{5\sqrt{2}}{2}) \).
If the normal at point \(t\) meets the parabola again at point \(t'\), the relationship between them is \(t' = -t - \frac{2}{t}\).
We can find the parameter \(t'\) for the point Q.
\(t' = -\sqrt{2} - \frac{2}{\sqrt{2}} = -\sqrt{2} - \sqrt{2} = -2\sqrt{2}\).
Now we find the coordinates of Q using the parameter \(t'\).
The x-coordinate of Q is \(a(t')^2 = \frac{5}{4}(-2\sqrt{2})^2 = \frac{5}{4}(4 \cdot 2) = 10\).
The y-coordinate of Q is \(2a(t') = 2(\frac{5}{4})(-2\sqrt{2}) = \frac{5}{2}(-2\sqrt{2}) = -5\sqrt{2}\).
So, the coordinates of Q are \( (10, -5\sqrt{2}) \).