Question

The point of intersection of normals to the parabola y^2=4x at the points whose ordinates are 4 and 6 is

Answer 1

Given :

Parabola  :- $y^{2} = 4x$

ordinates are 4 and 6

To Find :

The point of intersection of normals.

Solution :

If the ordinate is 4 then ,

$y^{2} = 4x$

substitute 4 in y

⇒ $4^{2} = 4x$

⇒ $x = 16$

Point for ordinate 4 is (4 , 4)

If the ordinate is 6 then,

Substitute 6 in y

⇒ $6^{2} =4x$

⇒ $36 = 4x$

⇒ $x = 9$

Point for ordinate 6 is (9, 6)

Now,

⇒ $y^{2} = 4x$

⇒ $2y\frac{dy}{dx} = 4$

⇒ $\frac{dy}{dx} = \frac{2}{y}$

Now, lets find the points of the normal

Normal at the point (4 , 4)

⇒ $y - 4 = - \frac{1}{(\frac{dy}{dx})_{(4,4)} } ( x - 4)$

⇒ $y - 4 = - \frac{1}{\frac{2}{4} } (x - 4 )$

⇒ $2x - y = 12 -------(1)$

Normal at the point (9 , 6)

⇒ $y - 6 = -\frac{1}{(\frac{dy}{dx})_{( 9 , 6 )} } (x - 9)$

⇒ $y - 6 = - \frac{1}{\frac{2}{6} } ( x - 9)$

⇒ $3x + y = 33 -------(2)$

Now by subtracting (2) from (1) we get

⇒ $3x + y = 33$

⇒ $2x + y = 12$

we get,

⇒ $- x = - 21$

⇒ $x = 21$

Now insert the x value in any of the above equation,

substitute x in equation (1)

⇒ $2x + y = 12$

⇒ $42 + y = 12$

⇒ $y = -30$

Therefore the point of intersection is ( 21 , -30).