Discussion Forum

 Let E₁  and E₂  be two ellipse whose centres are at the origin. The major axes of E₁ and E₂ lie along the lie X-axis and Y-axis, respectively. Let S be the circle   $x^{2}+(y-1)^{2}=2$ . The straight-line x+y=3 touches the curve S. E₁ and E₂ at P,Q and R, respectively. Suppose that  PQ=PR= $\frac{2\sqrt{2}}{3}$. If e₁ and e₂ are eccentricities  of E₁ and E₂ respectively, then the correct expression(s) is/are

Answer:

Explanation:

Here,    $E_{1}:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,(a>b)$

$E_{2}:\frac{x^{2}}{c^{2}}+\frac{y^{2}}{d^{2}}=1,(c<d)$

 and    $S:x^{2}+(y-1)^{2}=2$

 as tangent   to E₁ ,E₂ and S  is x+y=3.

 Let the point of contact of tangent be (x1,y1) to S.

$\therefore$  x.x₁+y.y₁-(y+y₁)+1=2

or   $xx_{1}+yy_{1}-y=(1+y_{1})$   , same as x+y=3.

 $\Rightarrow$         $\frac{x_{1}}{1}=\frac{y_{1}-1}{1}=\frac{1+y_{1}}{3}$

i.e,       x₁=1   and y₁=2

  $\therefore$  P= (1,2)

Since, PR= PQ=  $\frac{2\sqrt{2}}{3}$. thus   by parametric form.

   $\frac{x-1}{-1/\sqrt{2}}=\frac{y-2}{1/\sqrt{2}}=\pm\frac{2\sqrt{2}}{3}$

  $\Rightarrow $         $\left(x=\frac{5}{3},y=\frac{4}{3}\right)$

                                            and                  $ \left(x=\frac{1}{3},y=\frac{8}{3}\right)$

  $\therefore$       $ Q=\left(\frac{5}{3},\frac{4}{3}\right)$ and     $R= \left(\frac{1}{3},\frac{8}{3}\right)$

  Now, equation of tangent at Q on ellipse E₁  is

  $\frac{x.5}{a^{2}.3}+\frac{y.4}{b^{2}.3}=1$

   On comparing with x+y=3, we get

  a² =5  and b²=4

$\therefore$        $e_1^2=1-\frac{b^{2}}{a^{2}}=1-\frac{4}{5}=\frac{1}{5}$  ......(i)

 Also equation of tangent at R on ellipse E₂  is

   $\frac{x.1}{a^{3}.3}+\frac{y.8}{b^{2}.3}=1$

 On comparing with x+y=3, we get

    a²=1, b²=8

       $\therefore$        $e_2^2=1-\frac{a^{2}}{b^{2}}=1-\frac{1}{8}=\frac{7}{8}$  ......(ii)

 Now,   $e_1^2 .e_2^2=\frac{7}{40}\Rightarrow e_{1}e_{2}=\frac{\sqrt{7}}{2\sqrt{10}}$

and        $e_1^2+e_2^2=\frac{1}{5}+\frac{7}{8}=\frac{43}{40}$

          $|e_1^2+e_2^2|=|\frac{1}{5}-\frac{7}{8}|=\frac{27}{40}$