Question

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

A. 3x – y = 8
B. 3x + y + 8 = 0
C. x + 3y ± 8 = 0
D. x + 3y = 0

Answer 1

Given the equation of the line is 3x² – y² = 8

Now differentiating both sides with respect to x, we get

Now applying the sum rule of differentiation and differentiation of constant =0, so we get

So this is the slope of the given curve.

We know the slope of the normal to the curve is

Now given the equation of the line x + 3y = 8

⇒ 3y=8-x

Differentiation with respect to x, we get

Now as the normal to the curve is parallel to this line, hence the slope of the line should be equal to the slope of the normal to the given curve,

⇒ 3y=3x

⇒ y=x

On substituting this value of the given equation of the curve, we get

3x² – y² = 8

⇒ 3x² – (x)² = 8

⇒ 2x² = 8

⇒ x² = 4

⇒ x=± 2

When x=2, the equation of the curve becomes,

3x² – y² = 8

⇒ 3(2)² – y² = 8

⇒ 3(4) – y² = 8

⇒ 12-8= y²

⇒ y² = 4

⇒ y=±2

When x=-2, the equation of the curve becomes,

3x² – y² = 8

⇒ 3(-2)² – y² = 8

⇒ 3(4) – y² = 8

⇒ 12-8= y²

⇒ y² = 4

⇒ y=±2

So, the points at which normal is parallel to the given line are (±2, ±2).

And required equation of the normal to the curve at (±2, ±2) is

⇒ 3(y-(±2))=-(x-(±2))

⇒ 3y-(±6)=-x+(±2)

⇒ x+3y-(±6)-(±2)=0

⇒ x+3y+(± 8)=0

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

So the correct option is option C.