Mathematics - Coordinate Geometry Mixed Question with Solution | TestHub

(A) $x=y^2-6y+6\Rightarrow (y-3{)}^2=x+3$ : vertex $(-3,3)$, so the line joining the origin to the vertex has slope -1. Normal slope $=-\frac{dx}{dy}=-\left(2y_1-6\right)=-1\Rightarrow y_1=\frac{7}{2},x_1=\frac{49}{4}-21+6=-\frac{11}{4}$. Equation of normal: $y-\frac{7}{2}=-\left(x+\frac{11}{4}\right)\Rightarrow 4x+4y-3=0\to \mathrm{S}$. (B) For $xy=4$ : foci $(\pm 2\sqrt{2},\pm 2\sqrt{2})$; slope of latus rectum $=-1$ : $y-2\sqrt{2}=-(x-2\sqrt{2})\Rightarrow x+y=4\sqrt{2}\to \mathrm{P}$. (C) Reflection of (4, 3) in $x+y=3$ is (0, -1); reflected ray passes through (0, -1) and $(3,8):y+1=3x\Rightarrow y=3x-1\to \mathrm{Q}$. (D) Normal at $t_1:y=-t_{1}x+2t_1+t_1^3$ meets the parabola again at $t_2=-t_1-\frac{2}{t_1}$. The chord subtends a right angle at the origin: $t_1{t}_2=-4\Rightarrow -t_1^2-2=-4\Rightarrow t_1^2=2\Rightarrow t_1=-\sqrt{2}$ (slope of normal is positive). Equation: $y=\sqrt{2}x-2\sqrt{2}-2\sqrt{2}$, i.e. $\sqrt{2}y=2(x-4)\to \mathrm{R}$.