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