Question

If P₁ and P₂ are two points on the ellipse `x^2/4 + y^2` = 1 at which the tangents are parallel to the chord joining the points (0, 1) and (2, 0), then the distance between P₁ and P₂ is ______.

Options

• `2sqrt(2)`

• `sqrt(5)`

• `2sqrt(3)`

• `sqrt(10)`

Answer 1

If P₁ and P₂ are two points on the ellipse `x^2/4 + y^2` = 1 at which the tangents are parallel to the chord joining the points (0, 1) and (2, 0), then the distance between P₁ and P₂ is `underlinebb(sqrt(10))`.

Explanation:

Any tangent on an ellipse `x^2/4 + y^2` = 1 is given by y = `mx ± sqrt(a^2m^2 + b^2)`.

Here a = 2, b = 1

m = `(1 - 0)/(0 - 2)` = `-1/2`,

c = `sqrt(4(-1/2)^2 + 1^2)` = `sqrt(2)`

So, y = `-1/2 x ± sqrt(2)`

For ellipse : `x^2/4 + y^2/1` = 1

We put y = `-1/2 x + sqrt(2)`

∴ `x^2/4 + (-x/2 + sqrt(2))^2` = 1

`x^2/4 + (x^2/4 - 2(x/2) sqrt(2) + 2)` = 1

⇒ `x^2 - 2sqrt(2)x + 2` = 0

⇒ x = `sqrt(2)` or `-sqrt(2)`

If x = `sqrt(2)`, y = `1/sqrt(2)` and x = `-sqrt(2)`, y = `-1/sqrt(2)`

∴ Points are `(sqrt(2), 1/sqrt(2)), (-sqrt(2), -1/sqrt(2))`

∴ P₁P₂ = `sqrt({1/sqrt(2) - (-1/sqrt(2))}^2 + {sqrt(2) - (-sqrt(2))}^2`

= `sqrt((2/sqrt(2))^2 + (2sqrt(2))^2`

= `sqrt(2 + 8)`

= `sqrt(10)`