Question

The normal at the point (1, 1) on the curve 2y + x^ 2 = 3 is (A) x + y = 0 (B) x − y = 0 (C) x + y + 1 = 0 (D) x − y = 1

Answer 1

$curve - \\ {x}^{2} + 2y - 3 = 0 \\ diffrentiating \: wrt \:y \\ 2x \frac{dx}{dy} + 2 = 0 \\ \frac{dx}{dy} = - \frac{1}{x} \\ slope \: of \: normal = \frac{ - dx}{dy} = \frac{1}{x} \\ at \: point \: x = 1 \: it \: will \: be \: 1 \\ slope \: = 1 \\ m = \frac{y - y1}{x - x1} \\ 1 = \frac{y - 1}{x - 1} \\ x - y = 0$

Answer 2

we know, if y = f(x)
then, slope of normal of given curve = $\bf{-\frac{dx}{dy}}$

given, 2y + x² = 3
differentiate both sides,
2dy + 2x.dx = 0
=> x.dx = -dy
=> dx/dy = -1/x
so, slope of normal , m = -dx/dy = -(-1/x) = 1/x
at (1,1) slope of normal , m = $\bf{-\frac{dx}{dy}|_{(1,1)}}$=1/1 = 1
hence, m = 1

now, equation of normal :$\bf{y-y_1=m(x-x_1)}$
here $\bf{(x_1,y_1)}$ is point lies on the normal of curve and m is slope of normal.
e.g.,m = 1 and (x₁,y₁) = (1,1)
so, (y - 1) = 1(x - 1)
x - y = 0
hence, equation of normal is x - y = 0
therefore, option (B) is correct.