Exam

JEE Main

Year

2024

Date

9 April

Shift

Evening Shift

Subject

Mathematics

Chapter

Parabola

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Parabola

Let A, B and C be three points on the parabola y^2=6 x and let the...

Let $A, B and C$ be three points on the parabola $y^2=6 x$ and let the line segment $A B$ meet the line $L$ through $C$ parallel to the $x$-axis at the point $D$. Let $M and N$ respectively be the feet of the perpendiculars from $A and B$ on $L$. Then $\left(\frac{A M \cdot B N}{C D}\right)^2$ is equal to _________

Step-by-step solution

$$\begin{aligned} & \mathrm{m}_{\mathrm{AB}}=\mathrm{m}_{\mathrm{AD}} \\ & \Rightarrow \quad \frac{2}{\mathrm{t}_1+\mathrm{t}_2}=\frac{2 \mathrm{a}\left(\mathrm{t}_1-\mathrm{t}_3\right)}{\mathrm{at}_1^2-\alpha} \\ & \Rightarrow \quad \mathrm{at}_1^2-\alpha=\mathrm{a}\left\{\mathrm{t}_1^2-\mathrm{t}_1 \mathrm{t}_3+\mathrm{t}_1 \mathrm{t}_2-\mathrm{t}_2 \mathrm{t}_3\right\} \\ & \Rightarrow \quad \alpha=\mathrm{a}\left(\mathrm{t}_1 \mathrm{t}_3+\mathrm{t}_2 \mathrm{t}_3-\mathrm{t}_1 \mathrm{t}_2\right) \\ & \mathrm{AM}=\left|2 \mathrm{a}\left(\mathrm{t}_1-\mathrm{t}_3\right)\right|, \mathrm{BN}=\left|2 \mathrm{a}\left(\mathrm{t}_2-\mathrm{t}_3\right)\right|, \\ & \mathrm{CD}=\left|\mathrm{at}_3^2-\alpha\right|\end{aligned}$$$$\begin{aligned} & \mathrm{CD}=\left|\mathrm{at}_3^2-\mathrm{a}\left(\mathrm{t}_1 \mathrm{t}_3+\mathrm{t}_2 \mathrm{t}_3-\mathrm{t}_1 \mathrm{t}_2\right)\right| \\ &=\mathrm{a}\left|\mathrm{t}_3^2-\mathrm{t}_1 \mathrm{t}_3-\mathrm{t}_2 \mathrm{t}_3+\mathrm{t}_1 \mathrm{t}_2\right| \\ &=\mathrm{a}\left|\mathrm{t}_3\left(\mathrm{t}_3-\mathrm{t}_1\right)-\mathrm{t}_2\left(\mathrm{t}_3-\mathrm{t}_1\right)\right| \\ & \mathrm{CD}=\mathrm{a}\left|\left(\mathrm{t}_3-\mathrm{t}_2\right)\left(\mathrm{t}_3-\mathrm{t}_1\right)\right| \\ &\left(\frac{\mathrm{AM} \cdot \mathrm{BN}}{\mathrm{CD}}\right)^2=\left\{\frac{2 \mathrm{a}\left(\mathrm{t}_1-\mathrm{t}_3\right) \cdot 2 \mathrm{a}\left(\mathrm{t}_2-\mathrm{t}_3\right)}{\mathrm{a}\left(\mathrm{t}_3-\mathrm{t}_2\right)\left(\mathrm{t}_3-\mathrm{t}_1\right)}\right\}^2 \\ & 16 \mathrm{a}^2=16 \times \frac{9}{4}=36\end{aligned}$$$