Conic Sections Question 208
Question: The locus of mid point of that chord of parabola which subtends right angle on the vertex will be
[UPSEAT 1999]
Options:
A) $ y^{2}-2ax+8a^{2}=0 $
B) $ y^{2}=a(x-4a) $
C) $ y^{2}=4a(x-4a) $
D) $ y^{2}+3ax+4a^{2}=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Equation of parabola $ y^{2}=4ax $ ……(i) Equation of that chord of parabola whose mid point is $ (x_1,y_1) $ will be $ yy_1-2a(x+x_1)=y_1^{2}-4ax_1 $
or $ yy_1-2ax=y_1^{2}-2ax_1 $ or $ \frac{yy_1-2ax}{y_1^{2}-2ax_1}=1 $ …..(ii) Making equation (i) homogeneous by equation (ii), the equation of lines joining the vertex (0,0) of parabola to the point of intersection of chord (ii) and parabola (i) will be $ y^{2}=4ax\frac{yy_1-2ax}{y_1^{2}-2ax_1} $ or $ y^{2}(y_1^{2}-2ax_1)=4ax(yy_1-2ax) $
or $ 8a^{2}x^{2}-4ay_1xy+(y_1^{2}-2ax_1)y^{2}=0 $
If lines represented by it are mutually perpendicular, then coefficient of $ x^{2}+ $ coefficient of $ y^{2}=0 $
therefore, $ 8a^{2}+(y_1^{2}-2ax_1)=0 $ or $ y_1^{2}-2ax_1+8a^{2}=0 $ . Required locus of $ (x_1,y_1) $ is $ y^{2}-2ax+8a^{2}=0. $
BETA
