Question

If a tangent of slope $2$ of the ellipse $\left(\frac{x^2}{a^2}\right)+\left(\frac{y^2}{b^2}\right)=1$ is normal to the circle $x^2+y^2+4x+1=0$, then the maximum value of $ab$ is

• A. $4$
• B. $2$
• C. $1$
• D. None of these

Answer 1

The correct option is A

$4$

Step 1 : Solve for equation of tangent to ellipse with given slope

Tangent to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$with slope $m$ is given as

$y=mx\pm \sqrt{a^2{m}^2+b^2}$

substitute $m=2$ we get

$y=2x\pm \sqrt{4a^2+b^2}$

Step 2: Obtain co-ordinates of center of given circle

Compare given equation of circle with standard equation of circle $x^2+y^2+2gx+2hy+f=0$

We get $g=2,h=0$

Co-ordinates of center are given by $(-g,-h)$

Hence co-ordinates of center of given circle are $(-2,0)$

Step 3 : Apply condition for normal to circle

A normal to a circle always passes through the center.

The tangent to the ellipse is normal to the circle.

Hence the co-ordinates of the center satisfy the equation of the tangent

$\therefore$ $0=2(-2)\pm \sqrt{4a^2+b^2}$

$\Rightarrow$$4^2=4a^2+b^2$

$\Rightarrow$$16=4a^2+b^2$

Step 4: Solve for maximum possible value

$A.M\ge G.M$

$\Rightarrow$ $\frac{4a^2+b^2}{2}\ge \sqrt{4a^2{b}^2}$

$\Rightarrow$ $\frac{16}{2}\ge 2ab$

$\Rightarrow$ $ab\le 4$

Hence the maximum value possible for $ab$ is $4$.

Hence option (A) is the correct answer i.e. $4$