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Ellipse: Ellipse Centre Origin Lines Touch Ends Latus Rectum

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

Question
An ellipse has its centre at the origin, and the lines 3x+5y=203x + 5y = 20 and 3x−5y=203x - 5y = 20 touch it at the two ends of a latus rectum. Find the length of that latus rectum.
A0.160.16
B2.562.56
C3.23.2
D5.125.12correct
Solution
Question attachment Step 1: Adding the two given equations,
6x=40 ⟹ x=203,and then y=06x = 40 \ \Longrightarrow\ x = \frac{20}{3}, \qquad \text{and then } y = 0
 so the two tangents meet at (203, 0)\left(\dfrac{20}{3},\,0\right), a point of the xx-axis. Step 2: Tangents at the two ends of a latus rectum meet on the corresponding directrix, and by symmetry that meeting point lies on the major axis. Hence
ae=203.(1)\frac{a}{e} = \frac{20}{3} . \tag{1}
 Step 3: The tangent to x2a2+y2b2=1\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1 at the latus-rectum end (ae, b2a)\left(ae,\ \dfrac{b^{2}}{a}\right) is
x (ae)a2+y(b2/a)b2=1 ⟹ exa+ya=1 ⟹ ex+y=a\frac{x\,(ae)}{a^{2}} + \frac{y\left(b^{2}/a\right)}{b^{2}} = 1 \ \Longrightarrow\ \frac{ex}{a} + \frac{y}{a} = 1 \ \Longrightarrow\ ex + y = a
 Its xx-intercept is ae\dfrac{a}{e}, which is what Step 2 used, and its slope is −e-e. The given tangent 3x+5y=203x+5y=20 has slope −35-\dfrac35, so
e=35.(2)e = \frac35 . \tag{2}
 Step 4: From (1)(1) and (2)(2),
a=203⋅e=203⋅35=4a = \frac{20}{3}\cdot e = \frac{20}{3}\cdot\frac35 = 4
 Step 5:
b2=a2(1−e2)=16(1−925)=16⋅1625=25625b^{2} = a^{2}\left(1-e^{2}\right) = 16\left(1-\frac{9}{25}\right) = 16\cdot\frac{16}{25} = \frac{256}{25}
 Step 6:
ℓ=2b2a=2⋅256/254=512100=5.12\ell = \frac{2b^{2}}{a} = \frac{2\cdot 256/25}{4} = \frac{512}{100} = 5.12
 Answer: (4).
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