Question

Find the point on the ellipse \[4{x^2} + {y^2} = 4\] that are farthest

away from the point \(\left( {1,0} \right)\).

Short answer

The farthest points to the ellipse are \(\left( { - \frac{1}{3}, \pm \frac{4}{3}\sqrt 2 } \right)\).

Step-by-step solution

Application of Derivative

We can apply the derivative of a function to find the closest point of a curve to a point or the furthest point of a curve to another point. We find this point with the concept of local minimaand local maximarespectively.

We find local minima and maxima of a function by finding critical pointsand the second derivativeof the function.

Critical points are the \(x\) values that satisfies \(f'\left( x \right) = 0\).

The function has local maxima at a point \(x\) if \(f''\left( x \right) < 0\).

The function has a local minimum at a point \(x\) if \(f''\left( x \right) > 0\).

Finding the between the given point and a point of the curve

The equation of the ellipse is given by \(4{x^2} + {y^2} = 4\).

Let \(\left( {x,y} \right)\) be any point of the ellipse.

Now the distance between \(\left( {1,0} \right)\) and \(\left( {x,y} \right)\) will be

\(\begin{aligned}{l}d = \sqrt {{{\left( {x - 1} \right)}^2} + {{\left( {y - 0} \right)}^2}} \\d = \sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} \end{aligned}\)

Finding the Local Maxima of the square of the distance and hence the farthest point

Let \(S = {d^2} \Rightarrow S = {\left( {x - 1} \right)^2} + {y^2}\)

Using the ellipse equation we get

\(S = {\left( {x - 1} \right)^2} + \left( {4 - 4{x^2}} \right)\)

Now,

\(\begin{aligned}{l}S' = 2\left( {x - 1} \right) - 8x\\S' = - 6x - 2\end{aligned}\)

So the critical point will be

\(\begin{aligned}{c}S' = 0\\ - 6x - 2 = 0\\6x = - 2\\x = \frac{{ - 2}}{6}\\x = - \frac{1}{3}\end{aligned}\)

Hence the critical point is \(x = - \frac{1}{3}\).

The second derivative will be \(S'' = - 6 < 0\).

Hence the maximum value of the function \(S\) will be at the point \(x = - \frac{1}{3}\).

We see that \(S\left( { - 1} \right) = 4\),\(S\left( 1 \right) = 0\) and \(S\left( { - \frac{1}{3}} \right) = \frac{{16}}{3}\).

Hence \(\sqrt {\frac{{16}}{3}} \) is the maximum distance.

Also

\(\begin{aligned}{c}y = \pm \sqrt {4 - 4{{\left( { - \frac{1}{3}} \right)}^2}} \\ = \pm \sqrt {4 - \frac{4}{9}} \\ = \pm \sqrt {\frac{{32}}{9}} \\ = \pm \frac{4}{3}\sqrt 2 \end{aligned}\)

Hence the farthest points to the ellipse are \(\left( { - \frac{1}{3}, \pm \frac{4}{3}\sqrt 2 } \right)\).

Sketch of the ellipse