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Question
Prove that the curves y2 = 4x and x2 + y2 – 6x + 1 = 0 touch each other at the point (1, 2)
Solution
Given that the equation of the two curves are y2 = 4x .....(i)
And x2 + y2 – 6x + 1 = 0 .....(ii)
Differentiating (i) w.r.t. x, we get `2y "dy"/"dx"` = 4
⇒ `"dy"/"dx" = 2/y`
Slope of the tangent at (1, 2)
m1 = `2/2` = 1
Differentiating (ii) w.r.t. x
⇒ `2x + 2y * "dy"/"dx" - 6` = 0
⇒ `2y * "dy"/"dx"` = 6 – 2x
⇒ `"dy"/"dx" = (6 - 2x)/(2y)`
∴ Slope of the tangent at the same point (1, 2)
⇒ m2 = `(6 - 2 xx 1)/(2 xx 2)`
= `4/4`
= 1
We see that m1 = m2 = 1 at the point (1, 2).
Hence, the given circles touch each other at the same point (1, 2).
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