Let e_1 be the eccentricity of the hyperbola x^2 16- y^2 9=1 and e_2 be the eccentricity of the ellipse x^2 a^2+ y^2 b^2=1,a>b, which passes through…

The given equation of hyperbola is, $H:\frac{{x}^{2}}{16}-\frac{{y}^{2}}{9}=1$H:16x2​−9y2​=1

$$\Rightarrow {e}_{1}=\sqrt{1+\frac{{3}^{2}}{{4}^{2}}}=\sqrt{\frac{16+9}{16}}$$⇒e1​=1+4232​​=1616+9​​

$$\Rightarrow {e}_{1}=\frac{5}{4}$$⇒e1​=45​

Foci of this hyperbola is given by, $F(\pm \mathrm{ae},0)$F(±ae,0).

$$\Rightarrow F(\pm 5,0)$$⇒F(±5,0)

It is given that, ${e}_{1}{e}_{2}=1$e1​e2​=1

$$\Rightarrow {e}_{2}=\frac{4}{5}$$⇒e2​=54​

Also, ellipse is passing through $(\pm 5,0)$(±5,0)

$$\Rightarrow \frac{{5}^{2}}{{a}^{2}}+\frac{0}{{b}^{2}}=1$$⇒a252​+b20​=1

$$\Rightarrow \frac{25}{{a}^{2}}=1$$⇒a225​=1

$$\Rightarrow {a}^{2}=25$$⇒a2=25

$$\Rightarrow a=5$$⇒a=5

Now, ${e}_{2}=\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}$e2​=1−a2b2​​

$$\Rightarrow \frac{4}{5}=\sqrt{1-\frac{{b}^{2}}{25}}$$⇒54​=1−25b2​​

$$\Rightarrow \frac{16}{25}=1-\frac{{b}^{2}}{25}$$⇒2516​=1−25b2​

$$\Rightarrow \frac{{b}^{2}}{25}=\frac{9}{25}$$⇒25b2​=259​

$$\Rightarrow {b}^{2}=9$$⇒b2=9

$\therefore a=5$∴a=5 and $b=3$b=3

$$E:\frac{{x}^{2}}{25}+\frac{{y}^{2}}{9}=1$$E:25x2​+9y2​=1

Putting, $y=2$y=2

$$\Rightarrow \frac{{x}^{2}}{25}+\frac{4}{9}=1$$⇒25x2​+94​=1

$$\Rightarrow \frac{{x}^{2}}{25}=\frac{5}{9}$$⇒25x2​=95​

$$\Rightarrow {x}^{2}=\frac{125}{9}$$⇒x2=9125​

$$\Rightarrow x=\pm \frac{5\sqrt{5}}{3}$$⇒x=±355​​

So, end points of chord are $(\pm \frac{5\sqrt{5}}{3},2)$(±355​​,2)

Thus, length of chord $\mathrm{PQ}$PQ- is given by,

$${L}_{\mathrm{PQ}}=2\times \frac{5\sqrt{5}}{3}$$LPQ​=2×355​​

$$\Rightarrow {L}_{\mathrm{PQ}}=\frac{10\sqrt{5}}{3}$$⇒LPQ​=3105​​