Exam

JEE Main

Year

2024

Date

5 April

Shift

Morning Shift

Subject

Mathematics

Chapter

Straight Lines

JEE Main 2024 · 5 April · Morning Shift · Mathematics · Straight Lines

Let two straight lines drawn from the origin O intersect the line 3...

Let two straight lines drawn from the origin $\mathrm{O}$ intersect the line $3 x+4 y=12$ at the points $\mathrm{P}$ and $\mathrm{Q}$ such that $\triangle \mathrm{OPQ}$ is an isosceles triangle and $\angle \mathrm{POQ}=90^{\circ}$. If $l=\mathrm{OP}^2+\mathrm{PQ}^2+\mathrm{QO}^2$, then the greatest integer less than or equal to $l$ is :

Step-by-step solution

$$\begin{aligned} & 3 \mathrm{x}+4 \mathrm{y}=12 \\ & 3(\mathrm{r} \cos \theta)+4(\mathrm{r} \sin \theta)=12 \\ & \mathrm{r}(3 \cos \theta+4 \sin \theta)=12 \ldots(1) \\ & 3(-\mathrm{r} \sin \theta)+4(\mathrm{r} \cos \theta)=12 \\ & \mathrm{r}(-3 \sin \theta+4 \cos \theta)=12 \ldots(2) \\ & \left(\frac{12}{\mathrm{r}}\right)^2+\left(\frac{12}{\mathrm{r}}\right)^2=(3 \cos \theta+4 \sin \theta)^2+(-3 \sin \theta+4 \cos \theta)^2 \\ & 2\left(\frac{12}{\mathrm{r}}\right)^2=9+16 \\ & \frac{2 \times 144}{\mathrm{r}^2}=25 \Rightarrow 288=25 \mathrm{r}^2 \\ & \Rightarrow \frac{288}{25}=\mathrm{r}^2 \\ & \Rightarrow \sqrt{2}\left(\frac{12}{5}\right)=\mathrm{r} \\ & \ell=\mathrm{OP}^2+\mathrm{PQ}^2+\mathrm{QO}^2 \\ & \ell=\mathrm{r}^2+\mathrm{r}^2 \quad+\mathrm{r}^2(\cos \theta+\sin \theta)^2+\mathrm{r}^2(\sin \theta+\cos \theta)^2 \\ & =2 \mathrm{r}^2+\mathrm{r}^2(1+\sin 2 \theta+1-2 \sin 2 \theta) \\ & =2 \mathrm{r}^2+2 \mathrm{r}^2 \\ & =4 \mathrm{r}^2 \\ & =4\left(\frac{288}{25}\right)=\frac{1152}{25}=46.08 \\ & {[\ell]=46}\end{aligned}$$