Let a and b be any two numbers satisfying \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{4}. Then,   the   foot   of perpendicular from the origin on the variable line, \frac{x}{a}+\frac{y}{b}=1 , lies on :

 

  • Option 1)

    a hyperbola with each semi-axis =\sqrt{2}

  • Option 2)

     a hyperbola with each semi-axis = 2.

  • Option 3)

        a circle of radius = 2

  • Option 4)

       a circle of radius = \sqrt{2}

 

Answers (1)

As we learnt in

Equation of a line perpendicular to a given line -

Bx-Ay+\lambda =0  is the line perpendicular to Ax+By+C =0 .

 

- wherein

  \lambda is some other constant thanC.

 Equation of line perpendicular to AB

\frac{y-0}{x-0}=\frac{+a}{b}

ax - by = 0

Foot of perpendicular is 

\frac{x}{b}+\frac{y}{b}=1 and ax - by = 0

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

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

Also,   \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{4}

x^{2}+y^{2}= \frac{(a^{2}b)^{2}+(ab^{2})^{2}}{(a^{2}+b^{2})^{2}}= \frac{a^{2}b^{2}}{a^{2}+b^{2}}=4

Circle of radius is 2.


Option 1)

a hyperbola with each semi-axis =\sqrt{2}

This is incorrect option

Option 2)

 a hyperbola with each semi-axis = 2.

This is incorrect option

Option 3)

    a circle of radius = 2

This is correct option

Option 4)

   a circle of radius = \sqrt{2}

This is incorrect option

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