The equations of two sides of a variable tria | vedclass

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(C) The given parabola is $y^2 = 6x$. Comparing with $y^2 = 4ax$,we get $4a = 6$,so $a = \frac{3}{2}$.The equation of a tangent to the parabola $y^2 =

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$4y^2 - 18y + 3x + 18 = 0$

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(C) The given parabola is $y^2 = 6x$. Comparing with $y^2 = 4ax$,we get $4a = 6$,so $a = \frac{3}{2}$.The equation of a tangent to the parabola $y^2 = 6x$ with slope $m$ is $y = mx + \frac{a}{m} = mx + \frac{3}{2m}$.The triangle is formed by the lines $x = 0$,$y = 3$,and $y = mx + \frac{3}{2m}$.Finding the vertices of the triangle:$1$. Intersection of $x = 0$ and $y = 3$ is $(0, 3)$.$2$. Intersection of $x = 0$ and $y = mx + \frac{3}{2m}$ is $(0, \frac{3}{2m})$.$3$. Intersection of $y = 3$ and $y = mx + \frac{3}{2m}$ is $3 = mx + \frac{3}{2m} \Rightarrow mx = 3 - \frac{3}{2m} = \frac{6m - 3}{2m} \Rightarrow x = \frac{6m - 3}{2m^2}$. So the vertex is $(\frac{6m - 3}{2m^2}, 3)$.Let the circumcentre be $(h, k)$. Since the triangle is a right-angled triangle with vertices $(0, 3)$,$(0, \frac{3}{2m})$,and $(\frac{6m - 3}{2m^2}, 3)$,the circumcentre is the midpoint of the hypotenuse connecting $(0, \frac{3}{2m})$ and $(\frac{6m - 3}{2m^2}, 3)$.Thus,$h = \frac{0 + \frac{6m - 3}{2m^2}}{2} = \frac{6m - 3}{4m^2}$ and $k = \frac{3 + \frac{3}{2m}}{2} = \frac{6m + 3}{4m}$.From $k = \frac{6m + 3}{4m}$,we have $4mk = 6m + 3 \Rightarrow m(4k - 6) = 3 \Rightarrow m = \frac{3}{4k - 6} = \frac{3}{2(2k - 3)}$.Substitute $m$ into $h = \frac{6m - 3}{4m^2} = \frac{3(2m - 1)}{4m^2}$.After simplification,we get the locus $4y^2 - 18y + 3x + 18 = 0$.

The equations of two sides of a variable triangle are $x = 0$ and $y = 3$,and its third side is a tangent to the parabola $y^2 = 6x$. The locus of its circumcentre is:

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