The equation of the parabola whose focus is t | vedclass

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(C) Let the focus be $S(0, 0)$ and the tangent at the vertex be $x - y + 1 = 0$.Since the directrix is parallel to the tangent at the vertex,let the e

Vedclass · Accepted Answer

${x^2} + {y^2} + 2xy - 4x + 4y - 4 = 0$

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(C) Let the focus be $S(0, 0)$ and the tangent at the vertex be $x - y + 1 = 0$.Since the directrix is parallel to the tangent at the vertex,let the equation of the directrix be $x - y + \lambda = 0$.The distance from the focus $S(0, 0)$ to the tangent at the vertex is $d_1 = \frac{|0 - 0 + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{1}{\sqrt{2}}$.The distance from the focus $S(0, 0)$ to the directrix is $d_2 = \frac{|0 - 0 + \lambda|}{\sqrt{1^2 + (-1)^2}} = \frac{|\lambda|}{\sqrt{2}}$.Since the distance from the focus to the directrix is twice the distance from the focus to the tangent at the vertex,we have $d_2 = 2d_1$.$\frac{|\lambda|}{\sqrt{2}} = 2 	imes \frac{1}{\sqrt{2}} \implies |\lambda| = 2$.Since the focus $(0, 0)$ and the tangent at the vertex $x - y + 1 = 0$ are on opposite sides of the directrix,the directrix is $x - y + 2 = 0$.For any point $P(x, y)$ on the parabola,the distance to the focus equals the distance to the directrix: $SP^2 = PM^2$.$x^2 + y^2 = \left(\frac{|x - y + 2|}{\sqrt{1^2 + (-1)^2}}ight)^2$.$x^2 + y^2 = \frac{(x - y + 2)^2}{2}$.$2(x^2 + y^2) = x^2 + y^2 + 4 - 2xy + 4x - 4y$.$x^2 + y^2 + 2xy - 4x + 4y - 4 = 0$.

The equation of the parabola whose focus is the point $(0, 0)$ and the tangent at the vertex is $x - y + 1 = 0$ is

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