Conic Sections Test - 34

Given that,

Equations of diameter  $$3x-4y-7=0$$ ……  (1) and $$2x-3y-5=0$$……   (2)

Solve equations are

$$3x-4y-7=0$$     $$×2$$

$$2x-3y-5=0$$   $$×3$$

$$ 6x-8y-14=0 $$

$$ \underline{6x-9y-15=0}\,\,\,\,\,\,\,\,on\,\,subtracting\,\,\,that $$

$$ y+1=0 $$

$$ y=-1 $$

Put the value of in equation (1)

$$3x-4y-7=0$$ 

$$ 3x-4\left( -1 \right)-7=0 $$

$$ 3x+4-7=0 $$

$$ 3x-3=0 $$

$$ x=1 $$

Hence , the coordinates of the centre is $$(1,-1)$$

Also given area of circle= $$49\pi$$

$$ \pi {{r}^{2}}=49\pi  $$

$$ {{r}^{2}}=49 $$

$$ r=7 $$

Then, we know that the equation of circle is

$$ {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}} $$

$$ {{\left( x-1 \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{7}^{2}} $$

$$ {{x}^{2}}+1-2x+{{y}^{2}}+1+2y=49 $$

$$ {{x}^{2}}+{{y}^{2}}-2x+2y+2=49 $$

$$ {{x}^{2}}+{{y}^{2}}-2x+2y=49-2 $$

$$ {{x}^{2}}+{{y}^{2}}-2x+2y=47 $$

$$ {{x}^{2}}+{{y}^{2}}-2x+2y-47=0 $$

Hence, it is complete solution.

Option $$(C)$$ is correct answer.

We know that equation of ellipse is

  $$ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $$       …….(1)

Given that

  $$ e=\sqrt{\dfrac{2}{5}} $$

 $$ \sqrt{\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}}=\sqrt{\dfrac{2}{5}} $$

  $$ 5{{a}^{2}}-5{{b}^{2}}=2{{a}^{2}} $$

 $$ {{a}^{2}}=\dfrac{5{{b}^{2}}}{3} $$    …….(2)

$$\because $$ ellipse pass through (-3,1)

Then $$x=-3, y=1$$

Put in equation (1) we get

  $$ \dfrac{{{\left( -3 \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{1}^{2}}}{{{b}^{2}}}=1 $$

 $$ {{a}^{2}}+9{{b}^{2}}={{a}^{2}}{{b}^{2}} $$

 $$ \dfrac{5{{b}^{2}}}{3}+9{{b}^{2}}=\dfrac{5{{b}^{2}}}{3}.{{b}^{2}} $$

 $$ {{b}^{2}}=\dfrac{32}{5} $$    (From equation (1) and (2)  )

Put in equation (2) , we get $${{a}^{2}}=\dfrac{32}{3}$$

the value of a and b put in equation (1), we get

  $$ \dfrac{{{x}^{2}}}{\dfrac{32}{3}}+\dfrac{{{y}^{2}}}{\dfrac{32}{5}}=1 $$

 $$ 3{{x}^{2}}+5{{y}^{2}}=32 $$

This is required equation