- Home
- Standard 11
- Mathematics
10-2. Parabola, Ellipse, Hyperbola
hard
જો વર્તુળ $x^2+y^2-8 x=0$ અને અતિવલય $\frac{x^2}{9}-\frac{y^2}{4}=1$ ના છેદબિંદુઓ $A$ અને $B$ હોય તથા બિંદુ $P$ એ રેખા $2 x-3 y+4=0$ પ૨ ગતિ કરે, તો $\triangle PAB$ નું મધ્યકેન્દ્ર એ રેખા _________ પ૨ આવેલ છે.
A$4 x-9 y=12$
B$x+9 y=36$
C$9 x-9 y=32$
D$6 x-9 y=20$
(JEE MAIN-2025)
Solution
$x^2+y^2-8 x=0, \frac{x^2}{9}-\frac{y^2}{4}=1$
$4 x^2-9 y^2=36$
Solve $(1) \& (2)$
$4 x^2-9\left(8 x-x^2\right)=36$
$13 x^2-72 x-36=0$
$(13 x+6)(x=6)=0$
$ x =\frac{-6}{13}, x =6$
$x =\frac{-6}{13} \text { (rejected) }$
$y \rightarrow \text { Imaginary }$
$n =6, \frac{36}{9}-\frac{ y ^2}{4}=1$
$y ^2=12, y = I \sqrt{12}$
$A(6, \sqrt{12}), B (6,-\sqrt{12})$
$p \left(\alpha, \frac{2 \alpha+4}{3}\right) P$ lies on
$\text { centroid }( h , k )\ \ \ \ \ \ \ 2x – 3y + y = 0$
$h =\frac{12+\alpha}{3}, \alpha=3 h-12$
$k =\frac{\frac{2 \alpha+4}{3}}{3} \Rightarrow 2 \alpha+4=9 k$
$\alpha=\frac{9 k -4}{2}$
$6 h-2 y =9 k -4$
$6 x -9 y =20$
$4 x^2-9 y^2=36$
Solve $(1) \& (2)$
$4 x^2-9\left(8 x-x^2\right)=36$
$13 x^2-72 x-36=0$
$(13 x+6)(x=6)=0$
$ x =\frac{-6}{13}, x =6$
$x =\frac{-6}{13} \text { (rejected) }$
$y \rightarrow \text { Imaginary }$
$n =6, \frac{36}{9}-\frac{ y ^2}{4}=1$
$y ^2=12, y = I \sqrt{12}$
$A(6, \sqrt{12}), B (6,-\sqrt{12})$
$p \left(\alpha, \frac{2 \alpha+4}{3}\right) P$ lies on
$\text { centroid }( h , k )\ \ \ \ \ \ \ 2x – 3y + y = 0$
$h =\frac{12+\alpha}{3}, \alpha=3 h-12$
$k =\frac{\frac{2 \alpha+4}{3}}{3} \Rightarrow 2 \alpha+4=9 k$
$\alpha=\frac{9 k -4}{2}$
$6 h-2 y =9 k -4$
$6 x -9 y =20$
Standard 11
Mathematics
