Q

# The normal to the curve x^ 2 = 4y passing (1,2) is

23) The normal to the curve $x^2 = 4 y$ passing (1,2) is

(A) x + y = 3

(B) x – y = 3

(C) x + y = 1

(D) x – y = 1

Views

Given the equation of the curve
$x^2 = 4 y$
We know that the slope of the tangent at a point on the given curve is given by  $\frac{dy}{dx}$
$4\frac{dy}{dx} = 2x \\\ \frac{dy}{dx} = \frac{x}{2}$
We know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent } = \frac{-1}{\frac{x}{2}} = \frac{-2}{x}$
At point (a,b)
$Slope = \frac{-2}{a}$
Now, the equation of normal with point (a,b) and  $Slope = \frac{-2}{a}$

?
It is given that it also passes through the point (1,2)
Therefore,
$2-b=\frac{-2}{a}(1-a)\\ 2a -ba = 2a -2\\ ba = 2\\b =\frac{2}{a}$                  -(i)
It also satisfies equation $x^2 = 4 y\Rightarrow b = \frac{a^2}{4}$                    -(ii)
By comparing equation (i) and (ii)
$\frac{2}{a} = \frac{a^2}{4}\\ a^3 = 8\\ a = 2$
$b = \frac{2}{a} = \frac{2}{2} = 1$
$Slope = \frac{-2}{a} = \frac{-2}{2} = -1$

Now, equation of normal with point (2,1) and  slope = -1

$y-y_1=m(x-x_1)\\ y-1=-1(x-2)\\ y+x=3$