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

curve:-

In options equation of normal is given so,

Use hit and trial

A) x+y=3
put (x,y)= (1,2)

1+2=3=3
L.H.S =R.H.S ◆◆SATISFYING◆◆

B) x-y=3
put (x,y)=(1,2)

1-2=-1 ≠ 3

L.H.S ≠ R.H.S ●●NOT SATISFYING●●

C) x+y=1

put(x,y)=(1,2)

L.H.S
1+2=3

R.H.S=1

L.H.S ≠ R.H.S. ◆◆NOT SATISFYING◆◆

D)x-y=1

L.H.S
put (x,y)=(1,2)

=>1-2=-1

R.H.S=1

L.H.S ≠ R.H.S ◆◆◆NOT SATISFYING◆◆



so,
Option A is correct.

Answer 2

first of all we have to find out slope of normal of curve .
we know, if y = f(x) is given curve then,
slope of normal of curve = -$\bf{-\frac{dx}{dy}}$

given, x² = 4y
differentiate both sides,
2x.dx = 4dy
x.dx = 2dy
dx/dy = 2/x
so, slope of normal of curve , m = -dx/dy = -2/x
at (1,2) slope of normal of curve , m = $\bf{-\frac{dx}{dy}}|_{(1,2)}$ = -2/(2) = -1
hence, m = -1

now, equation of normal of curve :
$\bf{y-y_1=m(x-x_1)}$
here, $\bf{(x_1,y_1)}$ is the point lies on normal e.g., (1,2) and m is slope of normal of curve e.g.,m = -1

hence, (y -2) = -(x - 1)
y - 2 + x - 1 = 0
x + y = 3
hence, option (A) is correct