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
B
2x+y12a=0
C
2x+y18a=0
D
x+2y12a=0

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 answerMaths

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