JEE Main 2022 — Mathematics Coordinate Geometry

Given,

$$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1a>b$$a2x2​+b2y2​=1a>b

Now using eccentricity formula, ${e}^{2}=1-\frac{{b}^{2}}{{a}^{2}}$e2=1−a2b2​

We get, $\frac{1}{16}=1-\frac{{b}^{2}}{{a}^{2}}$161​=1−a2b2​

$$\Rightarrow \frac{{b}^{2}}{{a}^{2}}=1-\frac{1}{16}=\frac{15}{16}\Rightarrow {b}^{2}=\frac{15}{16}{a}^{2}$$⇒a2b2​=1−161​=1615​⇒b2=1615​a2

Now again $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$a2x2​+b2y2​=1 is passing through $(-4\sqrt{\frac{2}{5}},3)$(−452​​,3) on satisfying the point we get,

$$\Rightarrow \frac{16\times \frac{2}{5}}{{a}^{2}}+\frac{9}{{b}^{2}}=1$$⇒a216×52​​+b29​=1

$$\Rightarrow \frac{32}{5{a}^{2}}+\frac{9}{{b}^{2}}=1$$⇒5a232​+b29​=1

Now putting the value ${b}^{2}=\frac{15}{16}{a}^{2}$b2=1615​a2 in above equation we get,

$$\frac{32}{5{a}^{2}}+\frac{9}{\frac{15}{16}{a}^{2}}=1$$5a232​+1615​a29​=1

$$\frac{80}{5{a}^{2}}=1$$5a280​=1

$$16={a}^{2}$$16=a2

So, ${b}^{2}=15$b2=15 and ${a}^{2}+{b}^{2}=15+16=31$a2+b2=15+16=31