Study the following statements.I. The vertex | vedclass

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(C) For statement $I$: $x = ly^2 + my + n \Rightarrow x - n = l(y^2 + \frac{m}{l}y) \Rightarrow x - n + \frac{m^2}{4l} = l(y + \frac{m}{2l})^2 \Righta

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Statements $I$ & $III$ are true but $II$ is false

Vedclass · Answer

(C) For statement $I$: $x = ly^2 + my + n \Rightarrow x - n = l(y^2 + \frac{m}{l}y) \Rightarrow x - n + \frac{m^2}{4l} = l(y + \frac{m}{2l})^2 \Rightarrow (y + \frac{m}{2l})^2 = \frac{1}{l}(x - (n - \frac{m^2}{4l}))$. The vertex is $(n - \frac{m^2}{4l}, -\frac{m}{2l})$. Thus,statement $I$ is true.For statement $II$: $y = lx^2 + mx + n \Rightarrow y - n = l(x^2 + \frac{m}{l}x) \Rightarrow y - n + \frac{m^2}{4l} = l(x + \frac{m}{2l})^2 \Rightarrow (x + \frac{m}{2l})^2 = \frac{1}{l}(y - (n - \frac{m^2}{4l}))$. Here $4a = \frac{1}{l} \Rightarrow a = \frac{1}{4l}$. The focus is $(h, k+a) = (-\frac{m}{2l}, n - \frac{m^2}{4l} + \frac{1}{4l}) = (-\frac{m}{2l}, n - \frac{m^2-1}{4l})$. The given focus in the question is incorrect. Thus,statement $II$ is false.For statement $III$: The pole of $lx + my + n = 0$ with respect to $x^2 = 4ay$ is found by comparing with the chord of contact $xx_1 = 2a(y + y_1) \Rightarrow xx_1 - 2ay - 2ay_1 = 0$. Comparing coefficients: $\frac{x_1}{l} = \frac{-2a}{m} = \frac{-2ay_1}{n}$. Thus $x_1 = -\frac{2al}{m}$ and $y_1 = -\frac{n}{m}$. The given pole is $(\frac{-2al}{m}, \frac{n}{m})$,which is incorrect as the $y$-coordinate is $-\frac{n}{m}$. Thus,statement $III$ is false.

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