Points A,B,C lie on the parabola y² = 4ax . The tangents to the parabola at A,B,C taken in pairs intersect at points P,Q,R, then, the ratio of the areas of the $∆$ ABC and $∆$ PQR is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 2.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let $({at}_{1}^{2} 2 {at}_{1}), ({at}_{2}^{2} 2 {at}_{2}) and ({at}_{3}^{2} 2 {at}_{3})$ be the points A, B and C respectively.
$\begin{array}{l} ∴ Area of △ ABC = \frac{1}{2} | | \begin{array}{lll} {at}_{1}^{2} & 2 {at}_{1} & 1 \\ {at}_{2}^{2} & 2 {at}_{2} & 1 \\ {at}_{3}^{2} & 2 {at}_{3} & 1 \end{array} |= a^{2}| |\begin{array}{ccc} 1 & t_{1} & t_{1}^{2} \\ 1 & t_{2} & t_{2}^{2} \\ 1 & t_{3} & t_{3}^{2} \end{array}| ∥ \\ = a^{2} |(t_{1} - t_{2}) (t_{2} - t_{3}) (t_{3} - t_{1})| \end{array}$
Coordinates of P,Q and R are
$({at}_{2} t_{3}, a (t_{2} + t_{3})), ({at}_{3} t_{1}, a (t_{3} + t_{1})), ({at}_{1} t_{2}, a (t_{1} + t_{2}))$ respectively, then
$\begin{array}{l} ∴ Area of △ PQR = = |\begin{array}{lll} {at}_{2} t_{3} & a (t_{2} + t_{3}) & 1 \\ {at}_{3} t_{1} & a (t_{3} + t_{1}) & 1 \\ {at}_{1} t_{2} & a (t_{1} + t_{2}) & 1 \end{array}| = \frac{a^{2}}{2} |\begin{array}{lll} t_{2} t_{3} & t_{2} + t_{3} & 1 \\ t_{3} t_{1} & t_{3} + t_{1} & 1 \\ t_{1} t_{2} & t_{1} + t_{2} & 1 \end{array}| \\ = \frac{1}{2} \times Area of △ ABC \\ \frac{Area of △ ABC}{Area of △ PQR} = 2 \end{array}$