JEE Main 2022 — Mathematics Coordinate Geometry

By using eccentricty formula we get,

$$e=\sqrt{1+\frac{{b}^{2}}{{a}^{2}}},l=\frac{2{b}^{2}}{a}$$e=1+a2b2​​,l=a2b2​

Given ${e}^{2}=\frac{11}{14}l$e2=1411​l

So, $1+\frac{{b}^{2}}{{a}^{2}}=\frac{11}{14}\cdot \frac{2{b}^{2}}{a}$1+a2b2​=1411​⋅a2b2​

$$\frac{{a}^{2}+{b}^{2}}{{a}^{2}}=\frac{11}{7}\cdot \frac{{b}^{2}}{a}\cdots (1)$$a2a2+b2​=711​⋅ab2​⋯(1)

Also ${e}^{'}=\sqrt{1+\frac{{a}^{2}}{{b}^{2}}},{l}^{'}=\frac{2{a}^{2}}{b}$e′=1+b2a2​​,l′=b2a2​

Given ${({e}^{'})}^{2}=\frac{11}{8}{l}^{'}$(e′)2=811​l′

$$1+\frac{{a}^{2}}{{b}^{2}}=\frac{11}{8}\cdot \frac{2{a}^{2}}{b}$$1+b2a2​=811​⋅b2a2​

$$\frac{{a}^{2}+{b}^{2}}{{b}^{2}}=\frac{11}{4}\cdot \frac{{a}^{2}}{b}\cdots (2)$$b2a2+b2​=411​⋅ba2​⋯(2)

Now equation $(1)\div$(1)÷equation $(2)$(2) we get,

$$\frac{{b}^{2}}{{a}^{2}}=\frac{4}{7}\cdot \frac{{b}^{3}}{{a}^{3}}$$a2b2​=74​⋅a3b3​

$$\therefore 7a=4b\cdots (3)$$∴7a=4b⋯(3)

From $(2)$(2)

$$\frac{\frac{16{b}^{2}}{49}+{b}^{2}}{{b}^{2}}=\frac{11}{4}\cdot \frac{16{b}^{2}}{49b}$$b24916b2​+b2​=411​⋅49b16b2​

$$\therefore b=\frac{4\times 65}{11\times 16}\cdots (4)$$∴b=11×164×65​⋯(4)

We have to find value of

$$77a+44b$$77a+44b

So, $11(7a+4b)=11(4b+4b)=11\times 8b$11(7a+4b)=11(4b+4b)=11×8b

$\therefore$∴ Value of $11\times 8b=11\times 8\times \frac{4\times 65}{16\times 11}=130$11×8b=11×8×16×114×65​=130