Question

The normal at the point (1, 1) on the curve `2y + x^2` = 3 is

Options

• `x + y` = 0

• `(x - y)` = 0

• `x + y + 1` = 0

• `x - y` = 0

Answer 1

`(x - y)` = 0

Explanation:

The equation of the curve `2y + x^2` = 3, 

Differentiating,

`2 (dy)/(dx) + 2x = 0, (dy)/(dx) = - x`

= `(dy)/(dx)` at (1, 1) = – 1 = slope of tangent

Since slope of normal = `- 1/"Slope of tangent"` = 1

∴ The equation of normal is

∴ `y - y_1 = ("Slope of normal") (x - x_1)`

`y - 1 = 1(x - 1)` or `y - 1 = x - 1`

or `x - y` = 0