Find the equation of the normal to the curve x^2 = 4y which passes th…

The correct answer is Option (2) → Tangent: $x−y−1=0$; Normal: $x+y−3=0$

The given curve is $4y = x^2$   ...(i)

Let $P(x_1,y_1)$ be a point on the curve (i) the normal at which passes through the point (1, 2).

As $P(x_1,y_1)$ lies on the curve (i), $4y_1 = x_1^2$   ...(ii)

Diff. (i) w.r.t. x, we get $4\frac{dy}{dx}=2x⇒\frac{dy}{dx}=\frac{x}{2}$

∴ The slope of tangent at $P(x_1,y_1)=\frac{x_1}{2}$,

∴  the slope of normal at $P(x_1,y_1)=-\frac{2}{x_1}$.

∴  The equation of normal to the curve (i) at $P(x_1,y_1)$ is 

$y - y_1 = −\frac{2}{x_1} = (x − x_1)$   ...(iii)

Since it passes through the point (1, 2), we get

$2-y_1=-\frac{2}{x_1}(1 − x_1)⇒2-y_1=-\frac{2}{x_1}+2$

$⇒-y_1=-\frac{2}{x_1}⇒y_1=\frac{2}{x_1}$

From (ii) and (iv), we get $4.\frac{2}{x_1}=x_1^2⇒x_1^3=8⇒x_1=2$.

$∴y_1=\frac{2}{x_1}=\frac{2}{2}=1$

Thus, we get the point P(2, 1) on the curve (i), the normal at which passes through the point (1, 2).

Slope of the normal to curve (i) at P(2, 1) = $-\frac{2}{2}=-1$.

∴ The equation of normal is $y-1=-1(x-2)$ i.e. $x+y-3=0$.

Slope of tangent to curve (i) at $P = 1$.

∴ The equation of the corresponding tangent is

$y-1=1(x-2)$ i.e. $x-y-1 = 0$.