Answer 1

Given equation of the ellipse is x² + 4y² = 17.

∴ `x^2/17 + y^2/(17/4)` = 1

Comparing this equation with `x^2/"a"^2 + y^2/"b"^2` = 1 we get

a² = 17 and b² = `17/4`

Given equation of line is 8y + x = 17,

i.e., y = `(-1)/8 "x" + 17/8`

Comparing this equation with y = mx + c, we get

m = `(-1)/8` and c = `17/8`

For the line y = mx + c to be a tangent to the ellipse `x^2/"a"^2 + y^2/"b"^2` = 1, we must have

c² = a² m² + b²

c² = `(17/8)^7 = 289/64`

a²m² + b² = `17((-1)/8)^2 + 17/4`

= `17/64 + 17/4`

= `289/64`

= c²

∴ The given line touches the given ellipse and point of contact is

`((-"a"^2"m")/"c", "b"^2/"c") = ((-17((-1)/8))/(17/8), (17/4)/(17/8))`

= (1, 2).