Answer 1

We know that the equations of tangents with slope m to the ellipse `x^2/"a"^2 + y^2/"b"^2` = 1 are

y = `"m"x ± sqrt("a"^2"m"^2 + "b"^2)`   ...(1)

The equation of the ellipse is x² + 4y² = 9

∴ `x^2/9 + y^2/((9/4)` = 1

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

a² = 9, b² = `9/4`

Slope of 2x + 3y – 5 = 0 is `-2/3`

The required tangent is parallel to it

∴ its slope = m = `-2/3`

Using (1), the required equations of tangents are

y = `-(2x)/3 ± sqrt(9 xx 4/9 + 9/4)`

∴ y = `-(2x)/3 ± sqrt(25/4)`

∴ y = `-(2x)/3 ± 5/2`

∴ 6y = – 4x ± 15

∴ 4x + 6y = ± 15