Question

The point on the ellipse x² + 2y² = 6 closest to the line x + y = 7 is (a, b). The value of (a + b) will be ______.

विकल्प

• 0.00

• 1.00

• 2.00

• 3.00

Answer 1

The point on the ellipse x² + 2y² = 6 closest to the line x + y = 7 is (a, b). The value of (a + b) will be 3.00.

Explanation:

We have x² + 2y² = 6

⇒ `x^2/6 + y^2/3` = 1  ...(i)

Let P`(sqrt(6)cosθ, sqrt(3)sinθ)` be a point on (i) which is closest to the line x + y = 7.

Let PN be perpendicular to P on the line x + y = 7. PN is nearest (minimum) when it is along the normal at P.

The equation of the normal at P is `(sqrt(6)cosθ)x - (sqrt(3)  "cosec"  θ)y` = 3 it is perpendicular to x + y = 7

∴ `((sqrt(6)secθ)/(sqrt(3)  "cosec"  θ))x(-1)` = –1

⇒ tanθ = `1/sqrt(2)`

⇒ sinθ = `1/sqrt(3)`

and cosθ = `sqrt(2/3)`

∴ Required point is P`(sqrt(6) xx sqrt(2/3), sqrt(3) xx 1/sqrt(3)) ≡ (2, 1)`

Now, a + b = 2 + 1 = 3