Find the equation of the normal to the curve `x^2=4\ y` which passes through the point (1, 2).

Options

A. x+y = 3
B. x - y = 3
C. x+ y = 1
D. x- y = 1

Answer

A

Explanation

Given curve is,`x^(2) = 4y` …(1)
Let `(x_(1), y_(1))` be the point on curve (1) at which the normal drawn passes through the point(1, 2).
` x_(1) = 4y_(1)` ….(2)
Differentiate equation (1) w.r.t. X,
` 2x = 4(dy)/(dx) rArr (dy)/(dx) = (2x)/4 rArr (dy)/(dx) = x/2 `
` rArr ((dy)/(dx))_(x_(1), y_(1)) = x_(1)/2`
Now, equation of normal at point `(x_(1), x_(1))`
` :. y-y_(1) =- 1/((dy)/(dx)) _(x_(1), y_(1)) (x- x_(1))`
` rArr y - y_(1) = (-2)/x_(1) (x - x_(1))` ....(3)
but this normal passes through the point (1, 2)
` 2 - y_(1) =- 2/x_(1) (1 - x_(1) ) rArr 2 - y_(1) =- 2/x_(1)+ 2 `
` rArr y_(1) = 2 /x_(1)` ...(4)
From equations (2) and (4), ` x_(1)^(2) = 4 xx 2/x_(1)`
` rArr x_(1)^(3) = 8 rArr x_(1)= 2`
From equation (2), ` y_(1) = (x_(1)^(2))/4 = 2^(2)/4 = 1`
put ` x_(1)- 2 and y_(1) - 1 ` in equation (3), equation of normal
` y-1 =- 2/2 (x-2)`
` rArr y-1 =- x+2 rArr x+ y - 3 =0`
` rArr x+ y = 3` .