JEE Main 2017 — Mathematics Coordinate Geometry

Equation of ellipse whose centre at $(0,0)$(0,0) is

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

It passes through $(4,-1)$(4,−1)

$$\Rightarrow \frac{16}{{a}^{2}}+\frac{1}{{b}^{2}}=1$$⇒a216​+b21​=1

$$\Rightarrow 16{b}^{2}+{a}^{2}={a}^{2}{b}^{2}...(i)$$⇒16b2+a2=a2b2...(i)

Again it passes through $(-2,2)$(−2,2)

$$\Rightarrow \frac{4}{{a}^{2}}+\frac{4}{{b}^{2}}=1$$⇒a24​+b24​=1

$$\Rightarrow 4{b}^{2}+4{a}^{2}={a}^{2}{b}^{2}...(ii)$$⇒4b2+4a2=a2b2...(ii)

From the equation $(i)$(i) and $(ii)$(ii), we get,

$$16{b}^{2}+{a}^{2}=4{a}^{2}+4{b}^{2}$$16b2+a2=4a2+4b2

$$\Rightarrow {a}^{2}=4{b}^{2}$$⇒a2=4b2

Now, we have, ${e}^{2}=1-\frac{{b}^{2}}{{a}^{2}}=1-\frac{1}{4}=\frac{3}{4}$e2=1−a2b2​=1−41​=43​

$\Rightarrow e=\frac{\sqrt{3}}{2}$⇒e=23​​.