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Eccentricity of an Ellipse with Focus of xy = 1 and Asymptote as Directrix | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
Three curves are given:
C1:x2−y2=1 (x<0)C_1 : x^{2} - y^{2} = 1 \ (x < 0)
C2:xy=1 (y>0)C_2 : xy = 1 \ (y > 0)
 and C3C_3, an ellipse having one focus at a focus of C2C_2, the directrix corresponding to that focus being an asymptote of C1C_1, with the major axis of C3C_3 of length 83\dfrac83. Find the eccentricity of C3C_3.
A14\dfrac14
B122\dfrac{1}{2\sqrt2}
C12\dfrac{1}{\sqrt2}
D12\dfrac12correct
Solution
Question attachment Step 1: C2:xy=1C_2 : xy = 1 is a rectangular hyperbola. Turning the axes through 45∘45^{\circ} by
x=X−Y2,y=X+Y2x = \frac{X-Y}{\sqrt2}, \qquad y = \frac{X+Y}{\sqrt2}
xy=X2−Y22=1 ⟹ X22−Y22=1xy = \frac{X^{2}-Y^{2}}{2} = 1 \ \Longrightarrow\ \frac{X^{2}}{2} - \frac{Y^{2}}{2} = 1
 Step 2: a2=b2=2a^{2} = b^{2} = 2, so
e2=1+b2a2=2,e=2,ae=2e^{2} = 1 + \frac{b^{2}}{a^{2}} = 2, \qquad e = \sqrt2, \qquad ae = 2
 and the foci are X=±2X = \pm2, Y=0Y = 0. Step 3: Back in the original axes,
(x,y)=(X−Y2,X+Y2)=(±2,±2)(x,y) = \left(\frac{X-Y}{\sqrt2}, \frac{X+Y}{\sqrt2}\right) = \left(\pm\sqrt2, \pm\sqrt2\right)
 y>0y > 0 on C2C_2, so the focus of this branch is
F(2,2)F\left(\sqrt2, \sqrt2\right)
 Step 4: The asymptotes of C1:x2−y2=1C_1 : x^{2} - y^{2} = 1 are y=xy = x and y=−xy = -x. FF lies on y=xy = x, and a directrix cannot pass through its own focus, so the directrix of C3C_3 is
x+y=0x + y = 0
 Step 5: Distance from FF to this directrix,
∣2+2∣12+12=222=2\frac{\left|\sqrt2 + \sqrt2\right|}{\sqrt{1^{2}+1^{2}}} = \frac{2\sqrt2}{\sqrt2} = 2
 Step 6: For an ellipse the focus and its corresponding directrix are ae−ae\dfrac{a}{e} - ae apart. The major axis is 83\dfrac83, so a=43a = \dfrac43 and
ae−ae=2 ⟹ 43e−4e3=2\frac{a}{e} - ae = 2 \ \Longrightarrow\ \frac{4}{3e} - \frac{4e}{3} = 2
 Step 7: Multiplying by 3e3e (with e≠0e \ne 0),
4−4e2=6e ⟹ 2e2+3e−2=0 ⟹ (2e−1)(e+2)=04 - 4e^{2} = 6e \ \Longrightarrow\ 2e^{2} + 3e - 2 = 0 \ \Longrightarrow\ (2e-1)(e+2) = 0
 0<e<10 < e < 1 for an ellipse, so e=−2e = -2 is rejected and
e=12e = \frac12
 Step 8: a=43a = \dfrac43 and e=12e = \dfrac12 give ae=83\dfrac{a}{e} = \dfrac83 and ae=23ae = \dfrac23, and
83−23=2\frac83 - \frac23 = 2
 Answer: (4).
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