Question

Find equations of both tangent lines to the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\) that passes through the point (4,0).

Short answer

The equations of both tangent lines to the ellipse that passes through the point (4,0) are: \( y= -\frac{3}{2}(x-\frac{4}{\sqrt{5}})+\frac{6}{\sqrt{5}} \) and \( y= -\frac{3}{2}(x+\frac{4}{\sqrt{5}})-\frac{6}{\sqrt{5}} \).

Step-by-step solution

Identify the given ellipse and point

The given ellipse is \( \frac{x^{2}}{4}+\frac{y^{2}}{9}=1 \). The point through which the tangents pass is (4,0).

Differentiate the equation of an ellipse to get the slope

Rewrite the equation of the ellipse in the form \(y=\sqrt{9(1- \frac{x^{2}}{4})}\). Then, differentiate which results in \( y'= - \frac{9x}{2\sqrt{9(1- \frac{x^{2}}{4})}} \). This will provide the slope of the tangent.

Find the x-coordinates of tangent points

Set the derivative equal to the slope of the line passing through the point (4,0) and the point on the ellipse. Solve for x-values (x1, x2). This results in \( x_{1}= \frac{4}{\sqrt{5}}\) and \(x_{2}= -\frac{4}{\sqrt{5}} \).

Find the y-coordinates of tangent points

Substitute the x-values (x1, x2) into the equation of the ellipse and solve for y-value. This gives us two y-coordinates (y1, y2), where \(y_{1}= \frac{6}{\sqrt{5}}\), \(y_{2}= -\frac{6}{\sqrt{5}} \).

Form the equation of both tangent lines

Using point-slope form of equation, form the equations of two tangent lines passing through point (4,0) and (x1,y1) and (x2,y2) respectively. The equations are: \( y= \frac{-3}{2}(x-\frac{4}{\sqrt{5}})+\frac{6}{\sqrt{5}} \) and \( y= \frac{-3}{2}(x+\frac{4}{\sqrt{5}})-\frac{6}{\sqrt{5}} \).