Question

The equation of normal to the curve 3x² – y² = 8 which is parallel to the line x + 3y = 8 is ______.

पर्याय

• 3x – y = 8

• 3x + y + 8 = 0

• x + 3y ± 8 = 0

• x + 3y = 0

Answer 1

The equation of normal to the curve 3x² – y² = 8 which is parallel to the line x + 3y = 8 is x + 3y ± 8 = 0.

Explanation:

Given equation of the curve is 3x² – y² = 8   ......(i)

Differentiating both sides w.r.t. x, we get

`6x - 2y * "dy"/"dx"` = 0

⇒ `"dy"/"dx" = (3x)/y`

`(3x)/y` is the slope of the tangent

∴ Slope of the normal = `(-1)/("dy"/"dx") = (-y)/(3x)`

Now x + 3y = 8 is parallel to the normal

Differentiating both sides w.r.t. x, we have

`1 + 3 "dy"/"dx"` = 0

⇒ `"dy"/"dx" = - 1/3`

∴ `(-y)/(3x) = - 1/3`

⇒ y = x

Putting y = x in equation (i) we get

3x² – x² = 8

⇒ 2x² = 8

⇒ x² = 4

∴  x = ± 2 and y = ± 2

So the points are (2, 2) and (– 2, – 2).

Equation of normal to the given curve at (2, 2) is

y – 2 = `- 1/3(x - 2)`

⇒ 3y – 6 = – x + 2 

⇒ x + 3y – 8 = 0

Equation of normal at (– 2, – 2) is

y + 2 = `- 1/3 (x + 2)`

⇒ 3y + 6 = – x – 2

⇒ x + 3y + 8 = 0

∴ The equations of the normals to the curve are x + 3y ± 8 = 0.