JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 1)
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 :
$$4{y^2} - 18y - 3x - 18 = 0$$
$$4{y^2} + 18y + 3x + 18 = 0$$
$$4{y^2} - 18y + 3x + 18 = 0$$
$$4{y^2} - 18y - 3x + 18 = 0$$
Explanation
Third side of triangle
$t y=x+\frac{3}{2} t^{2}$
_25th_January_Evening_Shift_en_1_1.png)
$$ \begin{aligned} & \therefore H \equiv(0,3) \quad G \equiv\left(t-\frac{3 t}{2}\right) \end{aligned} $$
Let O (h, k)
_25th_January_Evening_Shift_en_1_2.png)
$$ \begin{aligned} & \Rightarrow \frac{2 h}{3}=t-\frac{t^{2}}{2} \& \frac{2 k+3}{3}=2+\frac{t}{2} \\\\ & \Rightarrow 4 h=6 t-3 t^{2} \& 4 k=6+3 t \\\\ & \Rightarrow 4 h=2(4 k-6)-3\left(\frac{(4 k-6)^{2}}{9}\right) \\\\ & \Rightarrow 4 h=6 k-9-\left(4 k^{2}+9-12 k\right) \\\\ & \Rightarrow 4 k^{2}-18 k+3 h+18=0 \\\\ & \Rightarrow 4 y^{2}-18 y+3 x+18=0 \end{aligned} $$
$t y=x+\frac{3}{2} t^{2}$
$$ \begin{aligned} & \therefore H \equiv(0,3) \quad G \equiv\left(t-\frac{3 t}{2}\right) \end{aligned} $$
Let O (h, k)
$$ \begin{aligned} & \Rightarrow \frac{2 h}{3}=t-\frac{t^{2}}{2} \& \frac{2 k+3}{3}=2+\frac{t}{2} \\\\ & \Rightarrow 4 h=6 t-3 t^{2} \& 4 k=6+3 t \\\\ & \Rightarrow 4 h=2(4 k-6)-3\left(\frac{(4 k-6)^{2}}{9}\right) \\\\ & \Rightarrow 4 h=6 k-9-\left(4 k^{2}+9-12 k\right) \\\\ & \Rightarrow 4 k^{2}-18 k+3 h+18=0 \\\\ & \Rightarrow 4 y^{2}-18 y+3 x+18=0 \end{aligned} $$
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