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