Question

If a tangent with slope 2 on the ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ is normal to the circle ${x^2} + {y^2} + 4x + 1 = 0$, then the maximum value of ab is.
$
  a.{\text{ }}4 \\
  b.{\text{ 2}} \\
  c.{\text{ 1}} \\
  d.{\text{ None of these}}{\text{.}} \\
$

Answer 1

Hint: Use the general equation of tangent to an ellipse which is Here m is the slope of the tangent to the ellipse. Later in the solution the basic definition of the arithmetic mean and geometric mean will help reach the maximum value of ab.

Complete step-by-step answer:

It is given that the slope of tangent of the ellipse is 2.
$ \Rightarrow m = 2.........\left( 1 \right)$
Now we know that the equation of tangent of an ellipse is $y = mx \pm \sqrt {{{\left( {ma} \right)}^2} + {b^2}} $, where m is the slope of the tangent.
Therefore from equation (1)
$y = 2x \pm \sqrt {{{\left( {2a} \right)}^2} + {b^2}} = 2x \pm \sqrt {4{a^2} + {b^2}} ............\left( 2 \right)$
Equation of circle is
${x^2} + {y^2} + 4x + 1 = 0$
Convert it into standard form.
So, add and subtract by 4 to make a complete square in x.
$
  {x^2} + 4 + 4x - 4 + {y^2} + 1 = 0 \\
   \Rightarrow {\left( {x + 2} \right)^2} + {y^2} = 3 \\
$
The standard equation of the circle is ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$, where (h, k) and r is the center and the radius of the circle respectively.
So, on comparing
$
   - h = 2,{\text{ }} - k = 0 \\
   \Rightarrow h = - 2,{\text{ k}} = 0 \\
$
So the center of the circle is (-2, 0).
Now it is given that the tangent is normal to the circle, therefore it passes through the center of the circle.
From equation (2)
$
  0 = 2\left( { - 2} \right) \pm \sqrt {4{a^2} + {b^2}} \\
   \Rightarrow 4 = \pm \sqrt {4{a^2} + {b^2}} \\
$
Now, squaring on both sides we have
\[
   \Rightarrow {4^2} = 4{a^2} + {b^2} \\
   \Rightarrow 4{a^2} + {b^2} = 16...........\left( 3 \right) \\
\]
Now consider two numbers $4{a^2}$ and ${b^2}$.
Arithmetic mean (A.M) of the numbers be $ = \dfrac{{4{a^2} + {b^2}}}{2}$
And the Geometric mean (G.M) of the numbers is $\sqrt {4{a^2}{b^2}} $
Now as we know that
$A.M \geqslant G.M$, so use this property
 $\therefore \dfrac{{4{a^2} + {b^2}}}{2} \geqslant \sqrt {4{a^2}{b^2}} $
Now from equation (3)
$
  \dfrac{{16}}{2} \geqslant \sqrt {4{a^2}{b^2}} \\
   \Rightarrow 8 \geqslant \sqrt {4{a^2}{b^2}} \\
$
Now, squaring on both sides we have
$
  64 \geqslant 4{a^2}{b^2} \\
   \Rightarrow {a^2}{b^2} \leqslant 16 \\
   \Rightarrow ab \leqslant \sqrt {16} \\
   \Rightarrow ab \leqslant 4 \\
$
Now as we see that the value of ab is less than or equal to 4.
$ \Rightarrow {\left( {ab} \right)_{\max }} = 4$
Hence option (a) is correct.

Note: In such types of questions always recall the equation of tangent which is stated above, then convert the equation of the circle into the standard form and find out the coordinates of the center then according to given condition substitute these coordinates into the equation of tangent, then always remember the property of A.M and G.M which is stated above, then simplify we will get the maximum value of ab.