To find the equation of the tangent to the ellipse \( 2x^2 + 9y^2 = 3 \) at the point \( (3, -1) \), follow these steps:
1. Verify that the point lies on the ellipse:
Substitute \( x = 3 \) and \( y = -1 \) into the ellipse equation: \[ 2(3)^2 + 9(-1)^2 = 2(9) + 9(1) = 18 + 9 = 27 \neq 3. \] The point \( (3, -1) \) does not lie on the ellipse \( 2x^2 + 9y^2 = 3 \). This means there is no tangent to the ellipse at this point.
However, if the ellipse equation is \( 2x^2 + 9y^2 = 27 \) (a corrected version), then the point \( (3, -1) \) lies on the ellipse: \[ 2(3)^2 + 9(-1)^2 = 2(9) + 9(1) = 18 + 9 = 27. \] 2. Equation of the tangent to an ellipse: The general equation of the tangent to the ellipse \( 2x^2 + 9y^2 = 27 \) at a point \( (x_1, y_1) \) is: \[ 2x_1 x + 9y_1 y = 27. \] 3. Substitute \( (x_1, y_1) = (3, -1) \): \[ 2(3)x + 9(-1)y = 27, \] \[ 6x - 9y = 27. \] 4. Simplify the equation:
Divide through by 3: \[ 2x - 3y = 9. \] Thus, the equation of the tangent to the ellipse \( 2x^2 + 9y^2 = 27 \) at the point \( (3, -1) \) is: \[ 2x - 3y = 9. \]