If a tangent of slope 2 of the ellipse x2-a2-y2-b2-1 is normal to the circle x2-y2-4 x-1-0 then the maximum value of ab is

The correct option is A 4A tangent of slope 2 is y=2x ±√(4a^2+b^2) ,this is normal to x^2+y^2+4x+1=0 then 0= 4 ±√(4a^2+b^2)⇒ 4a^2+b^2=16 using A.M.≥ G.M.,We ge ...

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Standard XII

Mathematics

Standard Equation of Ellipse

If a tangent ...

Question

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

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Solution

The correct option is B 4
A tangent of slope 2 is $y = 2 x \pm \sqrt{4 a^{2} + b^{2}}$ ,this is normal to $x^{2} + y^{2} + 4 x + 1 = 0$
then $0 = - 4 \pm \sqrt{4 a^{2} + b^{2}} \Rightarrow 4 a^{2} + b^{2} = 16$
$u s i n g A . M . \geq G . M ., W e g e t a b \leq 4$

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Similar questions

Q. If a tangent of slope 2 of the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

is normal to the circle

x^{2} + y^{2} + 4 x + 1 = 0

then the maximum value of ab is

Q. If a tangent to the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

having slope

2

is normal to the circle

x^{2} + y^{2} + 4 x + 1 = 0,

then the maximum value of

a b

Q. Slope of the common tangent to the ellipses

\frac{x^{2}}{a^{2} + b^{2}} + \frac{y^{2}}{b^{2}} = 1, \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} + b^{2}} = 1

is:

A

line touches the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

and the circle

x^{2} + y^{2} = r^{2}

Then the slope

m

of the common tangent is given by

m^{2} =

Q. Assertion :The angle of intersection between the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

and the circle

x^{2} + y^{2} = a b

{tan}^{- 1} \frac{(b - a)}{\sqrt{a b}}

Reason: The point of intersection of the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

and the circles

x^{2} + y^{2} = a b

⎛ ⎝ \sqrt{\frac{a^{2} b}{a + b}}, \sqrt{\frac{a b^{2}}{a + b}} ⎞ ⎠

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If a tangent of slope 2 of the ellipse x2a2+y2b2=1 is normal to the circle x2+y2+4x+1=0 then the maximum value of ab is

The correct option is B 4A tangent of slope 2 is y=2x±√4a2+b2 ,this is normal to x2+y2+4x+1=0 then 0=−4±√4a2+b2⇒4a2+b2=16 using A.M.≥G.M.,We get ab≤4

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

The correct option is B 4
A tangent of slope 2 is $y = 2 x \pm \sqrt{4 a^{2} + b^{2}}$ ,this is normal to $x^{2} + y^{2} + 4 x + 1 = 0$
then $0 = - 4 \pm \sqrt{4 a^{2} + b^{2}} \Rightarrow 4 a^{2} + b^{2} = 16$
$u s i n g A . M . \geq G . M ., W e g e t a b \leq 4$