1.

The equations of normal to the curve 3x2 - y2 = 8, such that it is parallel to the line x + 3y = 4, is

Answer»

Answer:

X+3Y+8=0

STEP-by-step explanation:

3x2−y2=8(1)

The GIVEN line is x+3y=4

consider the equation

3x2−y2=8

differentiate w.r.t x

we get

6x−2y dy/dx=0

⇒2y dy/dx=6x

dy/dx=3x/y

This is the slope of the tangent to the given curve.

∴ slope of the normal is −y3x

The slope of the given line is (−coefficientofx / coefficientofy)

i.e., −1/3

Step 2

Since the normal is parallel to the given line, the slopes are equal

∴−y/3x=−1/3

⇒y=x

Substituting for y

in equation (1) we get

3x2−x2=8

2x2=8

x2=4

∴x=±2

⇒y=±2

hence the POINTS of intersection are (±2,±2)

Step 3

Hence the equation of the normal is when (x,y)=(2,2)

(y−y1)=−1m(x−x1)

⇒(y−2)=−13(x−2)

3y−6=−x+2

⇒x+3y−8=0

Step 4

When (x,y)=(−2,−2)

(y+2)=−13(x+2)

⇒x+3y+8=0

Hence the equations of the normal are

x+3y±8=0

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