Question

Find the equation of the normal lines to the curve 3x² – y² = 8 which are parallel to the line x + 3y = 4.

Answer 1

We have equation of the curve 3x² – y² = 8

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

⇒ `6x - 2y * "dy"/"dx"` = 0

⇒ `-2y "dy"/"dx"` = – 6x

⇒ `"dy"/"dx" = (3x)/y`

Slope of the tangent to the given curve = `(3x)/y`

∴ Slope of the normal to the curve = `- 1/((3x)/y) = - y/(3x)`

Now differentiating both sides the given line x + 3y = 4

⇒ `1 + 3 * "dy"/"dx"` = 0

⇒ `"dy"/"dx" = - 1/3`

Since the normal to the curve is parallel to the given line x + 3y = 4.

∴ `- y/(3x) = - 1/3`

⇒ y = x

Putting the value of y in 3x² – y² = 8, we get

3x² – x² = 8

⇒ 2x² = 8

⇒ x² = 4

⇒ x = ± 2

∴ y = ± 2

∴ The points on the curve are (2, 2) and (– 2, – 2).

Now equation of the normal to the curve at (2, 2) is

y – 2 = `- 1/3 (x - 2)`

⇒ 3y – 6 = – x + 2 

⇒ x + 3y = 8

At (– 2, – 2) y + 2 = `- 1/3 (x + 2)`

⇒ 3y + 6 = – x – 2

⇒ x + 3y = – 8

Hence, the required equations are x + 3y = 8 and x + 3y = – 8 or x + 3y = ± 8.