The tangents at the points `(at_1^2,2at_1), (at_2^2, 2at_2)` on the parabola `y^2 = 4ax` are at right angles if

Options

A. `t_1t_2 =-1`
B. `t_1t_2=1`
C. `t_1t_2=2`
D. `t_1t_2=-2`

Answer

To determine the condition under which the tangents at the points \((at_1^2, 2at_1)\) and \((at_2^2, 2at_2)\) on the parabola \(y^2 = 4ax\) are at right angles, we can follow these steps:

### Step 1: Write the equations of the tangents
The equation of the tangent to the parabola \(y^2 = 4ax\) at the point \((at^2, 2at)\) is given by:
\[ yy_1 = 2a(x + x_1) \]
where \((x_1, y_1) = (at^2, 2at)\).

For the point \((at_1^2, 2at_1)\), the equation of the tangent is:
\[ yy_1 = 2a(x + at_1^2) \implies yy_1 = 2ax + 2aat_1^2 \]
This simplifies to:
\[ yy_1 - 2ax = 2aat_1^2 \]

For the point \((at_2^2, 2at_2)\), the equation of the tangent is:
\[ yy_2 = 2a(x + at_2^2) \implies yy_2 = 2ax + 2aat_2^2 \]
This simplifies to:
\[ yy_2 - 2ax = 2aat_2^2 \]

### Step 2: Find the slopes of the tangents
The slope of the tangent at the point \((at_1^2, 2at_1)\) is given by:
\[ m_1 = \frac{dy}{dx} = \frac{2a}{y} \text{ at } y = 2at_1 \implies m_1 = \frac{2a}{2at_1} = \frac{1}{t_1} \]

The slope of the tangent at the point \((at_2^2, 2at_2)\) is given by:
\[ m_2 = \frac{dy}{dx} = \frac{2a}{y} \text{ at } y = 2at_2 \implies m_2 = \frac{2a}{2at_2} = \frac{1}{t_2} \]

### Step 3: Condition for tangents to be perpendicular
For the tangents to be perpendicular, the product of their slopes must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \implies \frac{1}{t_1} \cdot \frac{1}{t_2} = -1 \]
This simplifies to:
\[ \frac{1}{t_1 t_2} = -1 \implies t_1 t_2 = -1 \]

### Conclusion
The condition for the tangents at the points \((at_1^2, 2at_1)\) and \((at_2^2, 2at_2)\) on the parabola \(y^2 = 4ax\) to be at right angles is:
\[ t_1 t_2 = -1 \]