Question

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

Answer 1

equations of normal : x + 3y ± 8 = 0

first of all, find slope of tangent using differentiation concept.

3x² - y² = 8

differentiating both sides with respect to x,

3(2x) - 2y dy/dx = 0

⇒dy/dx = 3x/y .......(1)

so, slope of normal = -1/slope of tangent

= -1/(3x/y)

= -y/3x

now slope of line (x + 3y = 8) is -1/3

as normal is parallel to line x + 3y = 8

so, slope of normal = slope of line

⇒-y/3x = -1/3

⇒y = x

then, 3x² - x² = 8

⇒2x² = 8

⇒x = ±2

there are two points , (2, -2) and (-2,-2)

equation of normal,

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

⇒3y - 6 + x - 2 = 0

⇒x + 3y - 8 = 0

now for (-2,-2), (y + 2) = -1/3(x + 2)

⇒x + 3y + 8 = 0

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Answer 2

Step-by-step explanation:

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