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Ellipse from a Minor Axis Line, Minor Axis Length and Foot of Directrix | JEE Advanced

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

Question
For an ellipse, the minor axis lies along the line x+2y=3x + 2y = 3 and has length 5\sqrt5, and the foot of a directrix is (2,3)(2,3). Then
Athe ellipse is 4(2x−y−1)2+2(x+2y−3)2=254(2x-y-1)^{2} + 2(x+2y-3)^{2} = 25correct
Bthe eccentricity is 12\dfrac{1}{\sqrt2}correct
Cthe major axis is 10\sqrt{10}correct
Dthe length of the latus rectum is 52\dfrac{\sqrt5}{\sqrt2}correct
Solution
Step 1: The major axis is perpendicular to the minor axis x+2y=3x+2y=3, whose slope is −12-\tfrac12, so its slope is 22:
2x−y=k2x - y = k
 Step 2: The directrix is perpendicular to the major axis and its foot lies on that axis, so (2,3)(2,3) satisfies this equation:
2(2)−3=1 ⟹ k=1,major axis: 2x−y=12(2) - 3 = 1 \ \Longrightarrow\ k = 1 , \qquad \text{major axis: } 2x - y = 1
 Step 3: From 2x−y=12x-y=1, y=2x−1y = 2x-1; substituting in x+2y=3x+2y=3,
x+2(2x−1)=3 ⟹ 5x=5 ⟹ x=1, y=1x + 2(2x-1) = 3 \ \Longrightarrow\ 5x = 5 \ \Longrightarrow\ x = 1,\ y = 1
 so the centre is C(1,1)C(1,1). Step 4: The centre-to-directrix distance is ae\dfrac{a}{e}:
ae=(2−1)2+(3−1)2=5.(1)\frac{a}{e} = \sqrt{(2-1)^{2} + (3-1)^{2}} = \sqrt5 . \tag{1}
 Step 5: 2b=52b = \sqrt5, so b=52b = \dfrac{\sqrt5}{2} and b2=54b^{2} = \dfrac54. From (1)(1), a=5 ea = \sqrt5\,e, and b2=a2(1−e2)b^{2} = a^{2}\left(1-e^{2}\right):
54=5e2(1−e2) ⟹ e2−e4=14 ⟹ (e2−12)2=0 ⟹ e=12.(2)\frac54 = 5e^{2}\left(1-e^{2}\right) \ \Longrightarrow\ e^{2} - e^{4} = \frac14 \ \Longrightarrow\ \left(e^{2}-\frac12\right)^{2} = 0 \ \Longrightarrow\ e = \frac{1}{\sqrt2} . \text{(2)}
 Step 6:
a=5⋅12=52,2a=252=10.(3)a = \sqrt5\cdot\frac{1}{\sqrt2} = \sqrt{\frac52}, \qquad 2a = 2\sqrt{\frac52} = \sqrt{10} . \text{(3)}
 a2=52>54=b2a^{2} = \dfrac52 > \dfrac54 = b^{2}, so the given axis is indeed the minor axis. Step 7: With a2=52a^{2} = \dfrac52,
ℓ=2b2a=2⋅545/2=5/25/2=52=52. ✓(D)\ell = \frac{2b^{2}}{a} = \frac{2\cdot\frac54}{\sqrt{5/2}} = \frac{5/2}{\sqrt{5/2}} = \sqrt{\frac52} = \frac{\sqrt5}{\sqrt2} . \ \checkmark \text{(D)}
 Step 8: The coordinate along the major axis is the distance from the minor-axis line, and the coordinate along the minor axis is the distance from the major-axis line, so
X=x+2y−35,Y=2x−y−15X = \frac{x+2y-3}{\sqrt5}, \qquad Y = \frac{2x-y-1}{\sqrt5}
 and X2a2+Y2b2=1\dfrac{X^{2}}{a^{2}} + \dfrac{Y^{2}}{b^{2}} = 1 becomes
(x+2y−3)25⋅52+(2x−y−1)25⋅54=1\frac{(x+2y-3)^{2}}{5\cdot\frac52} + \frac{(2x-y-1)^{2}}{5\cdot\frac54} = 1
2(x+2y−3)225+4(2x−y−1)225=1\frac{2(x+2y-3)^{2}}{25} + \frac{4(2x-y-1)^{2}}{25} = 1
4(2x−y−1)2+2(x+2y−3)2=25.(A)4(2x-y-1)^{2} + 2(x+2y-3)^{2} = 25 . \text{(A)}
 Answer: (1), (2), (3) and (4).
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