Question

Choose the correct answer in the following question:
The normal to the curve $x^2=4y$ passing through (1, 2) is

(a) x + y = 3 (b) x - y = 3 (c) x + y = 1 (d) x - y = 1

Answer 1

The given curve is $x^2=4y$ ...(i)
Let $(x_1,y_1)$ be a point on Eq. (i) at which normal passes through (1, 2), then $x_1^2=4y_1$ ...(ii)
On differentiating Eq. (i) w.r.t. x, we get
$2x=4\frac{dt}{dx}\Rightarrow \frac{dy}{dx}=\frac{x}{2}\Rightarrow (\frac{dy}{dx}{)}_{(x_1,y_1)}=\frac{x_1}{2}$
Now, equation of normal at $(x_1,y_1)$ is
$y-y_1=-{\frac{1}{\left(\frac{dy}{dx}\right)}}_{(x_1,y_1)}(x-x_1)\Rightarrow 2-y_1=-\frac{2}{x_1}(x-x_1)$ ...(iii)
But, this normal passes through (1, 2), therefore,
$2-y_1=-\frac{2}{x_1}(1-x_1)\Rightarrow 2-y_1=-\frac{2}{x_1}+2\Rightarrow y_1=\frac{2}{x_1}$ ...(iv)
From Eqs. (ii) and (iv), we get $x_1^2=4\times \frac{2}{x_1}\Rightarrow x_1^3=8\Rightarrow x_1=2$
Then, from Eq. (ii), $y_1=\frac{x_1^2}{4}=\frac{2^2}{4}=1$
The point on Eq. (i) at which the normal passes through the point (1, 2) is (2, 1). The required normal is obtained from Eq. (iii), by putting $x_1=2and{y}_1=1asy-1=-\frac{2}{2}(x-2)\Rightarrow y-1=-x+2\Rightarrow x+y-3=0$
Hence, the correct option is (a).