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Question
Answer the following:
A line touches the circle x2 + y2 = 2 and the parabola y2 = 8x. Show that its equation is y = ± (x + 2).
Solution
Given equation of the parabola is y2 = 8x
Comparing this equation with y2 = 4ax, we get
4a = 8
∴ a = 2
Equation of tangent to given parabola with slope m is y = `"m"x + 2/"m"`
∴ m2x – my + 2 = 0 ...(i)
Equation of the circle is x2 + y2 = 2
Its centre ≡ (0, 0) and Radius = `sqrt(2)`
Line (i) touches the circle.
∴ Length of perpendicular from the centre to the line (i) = radius
∴ `|("m"^2 0 - "m" 0 + 2)/sqrt("m"^4 + "m"^2)| = sqrt(2)`
∴ `4/("m"^4 + "m"^2)` = 2
∴ m4 + m2 – 2 = 0
∴ (m2 + 2)(m2 – 1) = 0
Since m2 ≠ – 2,
m2 – 1 = 0
∴ m = ± 1
When m = 1, equation of the tangent is
y = `(1)x + 2/((1))`
∴ y = (x + 2) ...(1)
When m = –1, equation of the tangent is
y = `(-1)x + 2/((-1))`
∴ y = –x – 2
∴ y = –(x + 2) ...(2)
From (1) and (2), the equation of the common tangents to the given parabola is y = ± (x + 2).
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