Question

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

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

Answer 1

Answer: (A)

$x+y=3$

Answer 2

The correct answer is A:$x+y=3$
The equation of the given curve is $x^2 = 4y$. Differentiating with respect to $x$, we have:
$2x=4.\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{x}{2}$
The slope of the normal to the given curve at point $(h, k)$ is given by,
$\frac{-1}{\frac{dy}{dx}]_{(h,k)}}=\frac{-2}{h}$
∴Equation of the normal at point $(h, k)$ is given as:
$y-k=\frac{-2}{h}(x-h)$
Now. it is given that the normal passes through the point $(1,2)$ Therefore, we have:
$2-k=\frac{-2}{h}(1-h) or k=2+\frac{2}{h}(1-h) ....(i)$
Since $(h, k)$ lies on the curve $x^2 = 4y$, we have $h^2 = 4k$
$k=\frac{h^2}{4}$
From equation (i), we have:
$\frac{h^2}{4}=2+\frac{2}{h}(1-h)$
$\frac{h^2}{4}=2h+2=2h=2$
$h^3=8$
$h=2$
$k=\frac{h^2}{4}=k=1$
Hence, the equation of the normal is given as:
$y=1=\frac{-2}{2}(x-2)$
$y-1=-(x-2)$
$x+y=3$
The correct answer is A.