Question

If a tangent of the slope \[\dfrac{1}{3}\] of the ellipse
\[\dfrac{{{x}^{3}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\left( a>b \right)\] is normal to the circle \[{{x}^{2}}+{{y}^{2}}+2x+2y+1=0\] then,

(a) the maximum value of ab is \[\dfrac{2}{3}\]
(b) \[a\in \left( \sqrt{\dfrac{2}{5}},2 \right)\]
(c) \[a\in \left( \dfrac{2}{5},2 \right)\]
(d) the maximum value of ab is 1

Answer 1

Hint: First of all write the equation of the tangent of the ellipse as \[y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\] and substitute in it the value of m and x and y as the center of the given circle. Now use \[AM\ge GM\] to find the maximum value of ab.

Complete step-by-step answer:

We are given that the tangent of the slope \[\dfrac{1}{3}\] of the ellipse \[\dfrac{{{x}^{3}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\left( a>b \right)\] is normal to the circle \[{{x}^{2}}+{{y}^{2}}+2x+2y+1=0\]. We have to check which option about the value of a or ab is correct. Let us consider the ellipse given in the question

\[\dfrac{{{x}^{3}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]

We know that the tangent of the above ellipse is given by \[y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\] where m is the slope of the ellipse. We are given that the slope of the ellipse is \[\dfrac{1}{3}\], so by substituting \[m=\dfrac{1}{3}\], we get,

\[y=\dfrac{1}{3}x\pm \sqrt{{{a}^{2}}{{\left( \dfrac{1}{3} \right)}^{2}}+{{b}^{2}}}\]

\[\Rightarrow y=\dfrac{1}{3}x\pm \sqrt{\dfrac{{{a}^{2}}}{9}+{{b}^{2}}}.....\left( i \right)\]

Now, we know that when a line is normal to the circle, it passes through its center. We are given that the tangent is normal to the circle \[{{x}^{2}}+{{y}^{2}}+2x+2y+1=0\]. So, it will pass through the center of the circle. By comparing the equation of the given circle by standard equation of the circle that is \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\], we get, g = 1, f = 1, and c =1. We know that the center of the circle is (– g, – f), so we get the center of the given circle as (– 1, – 1). Now by substituting x = – 1 and y = – 1 in equation (i), we get,

\[-1=\dfrac{-1}{3}\pm \sqrt{\dfrac{{{a}^{2}}}{9}+{{b}^{2}}}\]

\[\Rightarrow -1+\dfrac{1}{3}=\pm \sqrt{\dfrac{{{a}^{2}}}{9}+{{b}^{2}}}\]

\[\Rightarrow \dfrac{-2}{3}=\pm \sqrt{\dfrac{{{a}^{2}}}{9}+{{b}^{2}}}\]

By multiplying 9 on both the sides of the above equation, we get,

\[{{a}^{2}}+9{{b}^{2}}=4....\left( ii \right)\]

Now to find the approximate value of ab, we will apply

\[AM\ge GM....\left( iii \right)\]

We know that for two numbers, say p and q, \[AM=\dfrac{p+q}{2}\] and \[GM=\sqrt{pq}\].

So by taking two numbers as \[{{a}^{2}}\] and \[9{{b}^{2}}\], we get,

\[AM=\dfrac{{{a}^{2}}+9{{b}^{2}}}{2}\]

\[GM=\sqrt{{{a}^{2}}-9{{b}^{2}}}\]

By substituting the value of AM and GM in equation (iii), we get,

\[\dfrac{{{a}^{2}}+9{{b}^{2}}}{2}\ge \sqrt{{{a}^{2}}.9{{b}^{2}}}\]

\[\Rightarrow \dfrac{{{a}^{2}}+9{{b}^{2}}}{2}\ge 3ab\]

By substituting the value of \[{{a}^{2}}+9{{b}^{2}}\] from equation (ii), we get,

\[\dfrac{4}{2}\ge 3ab\]

\[\Rightarrow 3ab\le 2\]

\[ab\le \dfrac{2}{3}\]

Hence, we get the maximum value of ab as \[\dfrac{2}{3}\].

Therefore, option (a) is the right answer.

Note: In questions involving minima or maxima, students should always consider using this concept of \[AM\ge GM\]. Also, many students make the mistake of putting the wrong sign of greater than or lesser than while solving the question. So this must be taken care of. Also, remember that tangent of an ellipse is given as \[y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\] apart from deriving it every time and making the situation unnecessarily long.