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Midpoint of the Chord Cut on 9x - 4y = 13 by an Ellipse | JEE Advanced

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

Question
The line 9x−4y=139x - 4y = 13 cuts a chord on the ellipse x24+y29=1\dfrac{x^{2}}{4} + \dfrac{y^{2}}{9} = 1, and the mid point of that chord is (α,β)(\alpha, \beta). Find (α+β)(\alpha +\beta).
A00correct
B22
C11
D−2-2
Solution
Step 1: For a conic S=0S=0, the chord bisected at (x1,y1)\left(x_1,y_1\right) is T=S1T = S_1, so for x24+y29=1\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}=1,
xx14+yy19=x124+y129.(1)\frac{xx_1}{4} + \frac{yy_1}{9} = \frac{x_1^{2}}{4} + \frac{y_1^{2}}{9} . \tag{1}
 Step 2: (1)(1) and 9x−4y=139x - 4y = 13 are the same line, so their coefficients are proportional:
x1/49=y1/9−4\frac{x_1/4}{9} = \frac{y_1/9}{-4}
 Step 3:
x136=−y136 ⟹ y1=−x1\frac{x_1}{36} = -\frac{y_1}{36} \ \Longrightarrow\ y_1 = -x_1
 Step 4: With y1=−x1y_1 = -x_1, the right side of (1)(1) is
x124+x129=13x1236\frac{x_1^{2}}{4} + \frac{x_1^{2}}{9} = \frac{13x_1^{2}}{36}
 and the third ratio, against the constant term 1313, gives
x1/49=13x12/3613 ⟹ x136=x1236 ⟹ x12=x1 ⟹ x1=1 (x1≠0)\frac{x_1/4}{9} = \frac{13x_1^{2}/36}{13} \ \Longrightarrow\ \frac{x_1}{36} = \frac{x_1^{2}}{36} \ \Longrightarrow\ x_1^{2} = x_1 \ \Longrightarrow\ x_1 = 1 \ (x_1 \ne 0)
 x1=0x_1 = 0 would give y1=0y_1 = 0 and reduce (1)(1) to 0=00 = 0, which is not a line, so (α,β)=(1,−1)\left(\alpha,\beta\right) = (1,-1). Step 5:
14+19=1336<1.\frac{1}{4} + \frac{1}{9} = \frac{13}{36} < 1 .
 (1,−1)(1,-1) is inside the ellipse, so the chord is real. Step 6:
α+β=1+(−1)=0\alpha + \beta = 1 + (-1) = 0
 Answer: (1).
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