Question

Solve Q2 solve it full

Q.2. Find the coordinates of the focus, axis of the parabola, the equation of the directrix, the length of the latus rectum ends point of latus rectum :

$$\begin{gathered} \left(i\right)y^2=12x\left(ii\right)x^2=6y\left(iii\right)y^2=-8x\left(iv\right)x^2=-16y\left(v\right)y^2=10x \\ \left(vi\right)x^2=-9y \end{gathered}$$

Answer 1

Dear student
$$\begin{gathered} 1){\mathrm{y}}^2=12\mathrm{x} \\ \mathrm{On}\mathrm{comparing}\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{with}{\mathrm{y}}^2=4\mathrm{ax} \\ 4\mathrm{a}=12 \\ \Rightarrow \mathrm{a}=3 \\ \therefore \mathrm{Vertex}=(0,0) \\ \mathrm{Focus}=(\mathrm{a},0)=(3,0) \\ \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{directrix}: \\ \mathrm{x}=-\mathrm{a} \\ \Rightarrow \mathrm{x}=-3 \\ \mathrm{Axis}=\mathrm{y}=0 \\ \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{latus}\mathrm{ractum}=4\mathrm{a}=4\times 3=12\mathrm{units} \\ \mathrm{Endpoints}\mathrm{of}\mathrm{latus}\mathrm{ractum}=(\mathrm{a},\pm 2\mathrm{a})=\left(3,\pm 6\right) \\ 2){\mathrm{x}}^2=6\mathrm{y} \\ \mathrm{On}\mathrm{comparing}\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{with}{\mathrm{x}}^2=4\mathrm{ay} \\ 4\mathrm{a}=6 \\ \Rightarrow \mathrm{a}=\frac{3}{2} \\ \therefore \mathrm{Vertex}=(0,0) \\ \mathrm{Focus}=(0,\mathrm{a})=\left(0,\frac{3}{2}\right) \\ \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{directrix}: \\ \mathrm{y}=-\mathrm{a} \\ \Rightarrow \mathrm{y}=-\frac{3}{2} \\ \mathrm{Axis}=\mathrm{x}=0 \\ \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{latus}\mathrm{ractum}=4\mathrm{a}=4\times \frac{3}{2}=6\mathrm{units} \\ \mathrm{Endpoints}\mathrm{of}\mathrm{latus}\mathrm{ractum}=(\pm 2\mathrm{a},\mathrm{a})=\left(\pm 3,\frac{3}{2}\right) \end{gathered}$$
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