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Question
At what points on the curve x2 + y2 – 2x – 4y + 1 = 0, the tangents are parallel to the y-axis?
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Solution
Given that the equation of the curve is
x2 + y2 – 2x – 4y + 1 = 0 ....(i)
Differentiating both sides w.r.t. x, we have
`2x + 2y * "dy"/"dx" - 2 - 4 * "dy"/"dx"` = 0
⇒ `(2y - 4) "dy"/"dx"` = 2 – 2x
⇒ `"dy"/"dx" = (2 - 2x)/(2y - 4)` ....(ii)
Since the tangent to the curve is parallel to the y-axis.
∴ Slope `"dy"/"dx" = tan pi/2`
= `oo`
= `1/0`
So, from equation (ii) we get
`(2 - 2x)/(2y - 4) = 1/0`
⇒ 2y – 4 = 0
⇒ y = 2
Now putting the value of y in equation (i), we get
⇒ x2 + (2)2 – 2x – 8 + 1 = 0
⇒ x2 – 2x + 4 – 8 + 1 = 0
⇒ x2 – 2x – 3 = 0
⇒ x2 – 3x + x – 3 = 0
⇒ x(x – 3) + 1(x – 3) = 0
⇒ (x – 3)(x + 1) = 0
⇒ x = – 1 or 3
Hence, the required points are (– 1, 2) and (3, 2).
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