If the locus of a point that divides a chord | vedclass

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(A) Let the chord have slope $m = 2$. The equation of a chord with slope $m$ is $y = mx + c$,so $y = 2x + c$. Substituting into $y^2 = 4x$,we get $(2x

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$\left(\frac{2}{9}, \frac{8}{9}\right)$

Vedclass · Answer

(A) Let the chord have slope $m = 2$. The equation of a chord with slope $m$ is $y = mx + c$,so $y = 2x + c$. Substituting into $y^2 = 4x$,we get $(2x + c)^2 = 4x$,which simplifies to $4x^2 + (4c - 4)x + c^2 = 0$. Let the endpoints of the chord be $P(x_1, y_1)$ and $Q(x_2, y_2)$. By the section formula,the point $R(h, k)$ dividing $PQ$ in ratio $1:2$ is $h = \frac{1x_2 + 2x_1}{3}$ and $k = \frac{1y_2 + 2y_1}{3}$. Since $y_i = 2x_i + c$,we have $k = \frac{(2x_2 + c) + 2(2x_1 + c)}{3} = \frac{2(x_2 + 2x_1) + 3c}{3} = 2h + c$. Thus $c = k - 2h$. From the quadratic equation,$x_1 + x_2 = -\frac{4c - 4}{4} = 1 - c$. Then $3h = x_2 + 2x_1$. Also $x_1 + x_2 = 1 - c = 1 - (k - 2h) = 1 - k + 2h$. Substituting $x_2 = 3h - 2x_1$,we get $x_1 + 3h - 2x_1 = 1 - k + 2h$,so $x_1 = h + k - 1$. Since $x_1$ is a root of $4x^2 + (4c - 4)x + c^2 = 0$,we substitute $x_1$ and $c$: $4(h + k - 1)^2 + (4(k - 2h) - 4)(h + k - 1) + (k - 2h)^2 = 0$. Expanding this leads to the locus of $(h, k)$. After simplification,the vertex of the resulting parabola is $\left(\frac{2}{9}, \frac{8}{9}\right)$.

If the locus of a point that divides a chord of slope $2$ of the parabola $y^2 = 4x$ internally in the ratio $1:2$ is a parabola,then its vertex is

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