Consider the conic C: 25(x - 1)^2 + 25(y + 1) | vedclass

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(B) The given equation of the conic $C$ is $25((x - 1)^2 + (y + 1)^2) = (3x - 4y)^2$. Dividing by $25$,we get $(x - 1)^2 + (y + 1)^2 = \left(\frac{3x

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$2$

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(B) The given equation of the conic $C$ is $25((x - 1)^2 + (y + 1)^2) = (3x - 4y)^2$. Dividing by $25$,we get $(x - 1)^2 + (y + 1)^2 = \left(\frac{3x - 4y}{5}\right)^2$. This is in the form $SP^2 = PM^2$,where $S = (1, -1)$ is the focus and $3x - 4y = 0$ is the directrix. Since the eccentricity $e = 1$,the conic $C$ is a parabola. The locus of the point of intersection of perpendicular tangents to a parabola is its directrix. Thus,the curve $E$ is the line $3x - 4y = 0$. The minimum distance between the point $(2, -1)$ and the line $3x - 4y = 0$ is the perpendicular distance $d = \frac{|3(2) - 4(-1)|}{\sqrt{3^2 + (-4)^2}}$. $d = \frac{|6 + 4|}{\sqrt{9 + 16}} = \frac{10}{5} = 2$.

Consider the conic $C: 25(x - 1)^2 + 25(y + 1)^2 = (3x - 4y)^2$. If the curve $E$ is the locus of the point of intersection of perpendicular tangents to the conic $C$,then the minimum distance between the curve $E$ and the point $(2, -1)$ is:

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