Let A, B and C be three points on the parabola and let the line segment AB 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 is equal to __________.

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let A, B and C be three points on the parabola $y^{2} = 6 x$ and let the line segment AB 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 ${(\frac{A M \cdot B N}{C D})}^{2}$ is equal to __________. [2024]

Ans.
(36)

Equation of parabola, $y^{2} = 6 x$

$i . e ., y^{2} = 4 \times \frac{3}{2} x$

Let $A (\frac{3}{2} t_{1}^{2}, 3 t_{1}), B (\frac{3}{2} t_{2}^{2}, 3 t_{2}) and C (\frac{3}{2} t_{3}^{2}, 3 t_{3})$

be points on parabola $y^{2} = 6 x$ .

Equation of AB is given by

$\Rightarrow (y - 3 t_{1}) = \frac{3 t_{2} - 3 t_{1}}{\frac{3}{2} (t_{2}^{2} - t_{1}^{2})} (x - \frac{3}{2} t_{1}^{2})$

$\Rightarrow y - 3 t_{1} = \frac{2}{t_{1} + t_{2}} \frac{(2 x - 3 t_{1}^{2})}{2}$

$\Rightarrow (y - 3 t_{1}) (t_{1} + t_{2}) = 2 x - 3 t_{1}^{2}$

$\Rightarrow y (t_{1} + t_{2}) - 3 t_{1}^{2} - 3 t_{1} t_{2} = 2 x - 3 t_{1}^{2}$

$\Rightarrow y (t_{1} + t_{2}) = 2 x + 3 t_{1} t_{2}$

For point D, $y = 3 t_{3}$

$\Rightarrow 2 x = 3 t_{1} t_{3} + 3 t_{2} t_{3} - 3 t_{1} t_{2}$

$\Rightarrow x = \frac{3}{2} (t_{1} t_{3} + t_{2} t_{3} - t_{1} t_{2})$

$| C D | = \frac{3}{2} (t_{1} t_{3} + t_{2} t_{3} - t_{1} t_{2}) - \frac{3}{2} t_{3}^{2}$

$| A M | = | 3 t_{1} - 3 t_{3} |, | B N | = | 3 t_{2} - 3 t_{3} |$

So, ${(\frac{A M \cdot B N}{C D})}^{2} = {(\frac{9 (t_{1} - t_{3}) (t_{2} - t_{3})}{\frac{3}{2} (t_{1} - t_{3}) (t_{3} - t_{2})})}^{2} = {(- 6)}^{2} = 36$ .

Want Us to Email you About Special Offers & Updates?