The normal drawn at a point `(at_1^2,2at_1)1 )` on the parabola `y^2=4ax` meets it again at the point `(at_2^2,2at_2)`, then

Options

A. `t_1=2t_2`
B. `t_1^2=2t_2`
C. `t_1t_2=-1`
D. None of those

Answer

To solve the problem, we need to find the relationship between the points on the parabola \( y^2 = 4ax \) and the normal drawn at a point \( (at_1^2, 2at_1) \).

### Step 1: Identify the point on the parabola
The point on the parabola is given as \( (at_1^2, 2at_1) \).

### Step 2: Find the slope of the tangent at this point
The equation of the parabola is \( y^2 = 4ax \). To find the slope of the tangent at the point \( (at_1^2, 2at_1) \), we differentiate the equation of the parabola with respect to \( x \):

\[ \frac{dy}{dx} = \frac{2a}{y} = \frac{2a}{2at_1} = \frac{a}{at_1} = \frac{1}{t_1} \]

### Step 3: Find the slope of the normal
The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore, the slope of the normal is:

\[ -\frac{1}{\frac{1}{t_1}} = -t_1 \]

### Step 4: Write the equation of the normal
Using the point-slope form of the equation of a line, the equation of the normal at the point \( (at_1^2, 2at_1) \) is:

\[ y - 2at_1 = -t_1(x - at_1^2) \]

Simplifying this, we get:

\[ y - 2at_1 = -t_1x + at_1^3 \]
\[ y = -t_1x + at_1^3 + 2at_1 \]

### Step 5: Find the intersection of the normal with the parabola
We need to find where this normal intersects the parabola again. Substitute \( y = -t_1x + at_1^3 + 2at_1 \) into the parabola's equation \( y^2 = 4ax \):

\[ (-t_1x + at_1^3 + 2at_1)^2 = 4ax \]

Expanding this and rearranging will give us a quadratic equation in terms of \( x \).

### Step 6: Solve the quadratic equation
After substituting and simplifying, we will arrive at a quadratic equation of the form:

\[ Ax^2 + Bx + C = 0 \]

Using the quadratic formula \( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \), we can find the values of \( x \) where the normal intersects the parabola.

### Step 7: Find the corresponding \( y \) values
Once we have the \( x \) values, we can substitute them back into the equation of the normal or the parabola to find the corresponding \( y \) values.

### Conclusion
We will find that the normal intersects the parabola at two points: the original point \( (at_1^2, 2at_1) \) and another point \( (at_2^2, 2at_2) \). The relationship between \( t_1 \) and \( t_2 \) can be derived from the quadratic equation.