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Question

If $$P$$ is a point on the parabola $${ y }^{ 2 }=4x$$ in which the abscissa is equal to ordinate then the equation of the normal at $$P$$ is


A
2x+y+12a=0
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B
2x+y12a=0
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C
2x+y18a=0
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D
x+2y12a=0
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Solution

The correct option is B $$2x+y-12a=0$$
Since $$p$$ line on parabola and its coordinates satisfies
$$y = x$$  
putting $$y=x$$ in the equation
$$\begin{array}{l} { y^{ 2 } }=4x \\ { x^{ 2 } }=4x \\ { x^{ 2 } }-4x=0 \\ x\left( { x-4 } \right) =0 \\ x=0,4 \\ so \\ y=0,4 \end{array}$$
therefore point $$p$$ is $$\left( {0,0} \right)$$ or $$\left( {4,4} \right)$$
Now
$$\begin{array}{l} { y^{ 2 } }=4x \\ 2y\dfrac { { dy } }{ { dx } } =4 \\ \dfrac { { dy } }{ { dx } } =\dfrac { 2 }{ y }  \\ \left( { \dfrac { { dy } }{ { dx } }  } \right) \left( { 0,0 } \right) =\infty  \\ \left( { \dfrac { { dy } }{ { dx } }  } \right) \left( { 4,4 } \right) =\dfrac { 2 }{ 4 } =\dfrac { 1 }{ 2 }  \end{array}$$

equation of normal at $$\left( {0,0,} \right)$$ is
$$\begin{array}{l} y-0=\dfrac { { -1 } }{ { { { \left( { \dfrac { { dy } }{ { dx } }  } \right)  }_{ \left( { 0,0, } \right)  } } } } \left( { x-0 } \right)  \\ y=\dfrac { { -1 } }{ \infty  } x \\ y=0 \end{array}$$

equation of normal at $$\left( {4,4} \right)$$ is
$$\begin{array}{l} y-4=\dfrac { 1 }{ { { { \left( { \dfrac { { dy } }{ { dx } }  } \right)  }_{ \left( { 4,4 } \right)  } } } } \left( { x-4 } \right)  \\ y-4=\dfrac { { -1 } }{ { \dfrac { 1 }{ 2 }  } } \left( { x-4 } \right)  \\ y-4=-2\left( { x-4 } \right)  \\ y-4=-2x+8 \\ 2x+y-12=0 \end{array}$$
thus the required equation of normal is $${2x + y - 12a = 0}$$
$$B$$ is the correct answer

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