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The line y=sqrt(2)x+4sqrt(2) is normal t...

The line `y=sqrt(2)x+4sqrt(2)` is normal to `y^(2)=4ax` then a=

A

12

B

-12

C

36

D

-36

Text Solution

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The correct Answer is:
To solve the problem, we need to find the value of \( a \) for which the line \( y = \sqrt{2}x + 4\sqrt{2} \) is normal to the parabola \( y^2 = 4ax \). ### Step-by-Step Solution: 1. **Identify the equation of the normal to the parabola**: The general form of the equation of the normal to the parabola \( y^2 = 4ax \) is given by: \[ y = mx - 2am - am^3 \] where \( m \) is the slope of the normal line. 2. **Identify the slope of the given line**: The given line is: \[ y = \sqrt{2}x + 4\sqrt{2} \] From this, we can see that the slope \( m \) is \( \sqrt{2} \). 3. **Set up the equation**: We can now set up the equation using the slope we identified: \[ y = \sqrt{2}x - 2a(\sqrt{2}) - a(\sqrt{2})^3 \] This simplifies to: \[ y = \sqrt{2}x - 2\sqrt{2}a - 2a\sqrt{2} \] which can be rewritten as: \[ y = \sqrt{2}x - 4a\sqrt{2} \] 4. **Compare the two equations**: Now we compare this with the given line: \[ \sqrt{2}x + 4\sqrt{2} \] By comparing the constant terms, we have: \[ -4a\sqrt{2} = 4\sqrt{2} \] 5. **Solve for \( a \)**: To find \( a \), we can divide both sides by \( \sqrt{2} \): \[ -4a = 4 \] Dividing both sides by -4 gives: \[ a = -1 \] ### Final Answer: Thus, the value of \( a \) is: \[ \boxed{-1} \]
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