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Value of λ for (5x-1)²+(5y-2)²=λ(3x-4y-1)² Parabola | JEE

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

Question
If any point P(x,y)P(x, y) which satisfies the relation (5x−1)2+(5y−2)2=λ(3x−4y−1)2(5x - 1)^2 + (5y - 2)^2 = \lambda(3x - 4y - 1)^2 represents a parabola, then:
Aλ=1\lambda = 1correct
Bλ<1\lambda < 1
Cλ>1\lambda > 1
Dλ>2\lambda > 2
Solution
Step 1: Take 2525 out of each squared term and divide through.
(x−15)2+(y−25)2=λ(3x−4y−15)2.\left(x - \frac{1}{5}\right)^2 + \left(y - \frac{2}{5}\right)^2 = \lambda\left(\frac{3x - 4y - 1}{5}\right)^2.
 The left side is SP2SP^2 for the focus S(15,25)S\left(\dfrac{1}{5}, \dfrac{2}{5}\right), and since 32+42=5\sqrt{3^2 + 4^2} = 5, the right side is λ PM2\lambda\,PM^2 where PMPM is the distance from PP to 3x−4y−1=03x - 4y - 1 = 0. So SP2=λ PM2SP^2 = \lambda\,PM^2. Step 2: A conic obeys SP=e PMSP = e\,PM, and a parabola has e=1e = 1, hence SP2=PM2SP^2 = PM^2 and λ=1\lambda = 1. Correct answer: (1)
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