Question

If $ a\ne 0 $ and the line $ 2bx+3cy+4d=0 $ passes through the points of intersection of the parabolas $ {{y}^{2}}=4ax $ and $ {{x}^{2}}=4ay, $ then

• A. $ {{d}^{2}}+{{(2b+3c)}^{2}}=0 $
• B. $ {{d}^{2}}+{{(3b+2c)}^{2}}=0 $
• C. $ {{d}^{2}}+{{(2b-3c)}^{2}}=0 $
• D. $ {{d}^{2}}+{{(3b-2c)}^{2}}=0 $

Answer 1

Answer: (A)

The equation of parabolas are $ {{y}^{2}}=4ax $ and $ {{x}^{2}}=4ay $ . On solving these, we get $ x=0 $ and $ x=4a $ Also $ y=0 $ and $ y=4a. $ $ \therefore $ The point of intersection of parabolas are A(0,0) and $ B(4a,4a) $ . Also line $ 2bx+3cy+4d=0 $ passes through A and B. $ \therefore $ $ d=0 $ ...(i) and $ 2b.4a+3c-4a+4d=0 $ $ \Rightarrow $ $ 2ab+3ac+d=0 $ $ \Rightarrow $ $ a(2b+3c)=0 $ $ (\because d=0) $ $ \Rightarrow $ $ 2b+3c=0 $ ...(ii) On squaring Eqs. (i) and (ii) and then adding, we get $ {{d}^{2}}+{{(2b+3c)}^{2}}=0 $