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).

Answer 1

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).