Explanation:
C Given, equation of parabola,
\(y^2=4 a x\) and \(x^2=4 a y\)
\(y=\frac{x^2}{4 a}\)
Substitute the value of \(y\),
\(\frac{x^4}{16 a^2}=4 a x\)
\(x^4-64 a^3 x=0\)
\(\Rightarrow x\left(x^3-64 a^3\right)=0\)
\(x=0,4 a\)
For, \(x=0,4 a, y=0, y=4 a\)
The point of intersection of parabola are \((0,0)\) and \((4 a\), 4a)
Now the given line \(2 b x+3 c y+4 d=0\) passes through \((0,0)\) and \((4 a, 4 a)\)
\(\mathrm{d}=0\)
And
\(2 b \cdot 4 a+3 c \cdot 4 a+4 d=0\)
\(2 a b+3 a c+d=0\)
\(a(2 b+3 c)=0\)
\(2 b+3 c=0\)
On squaring and adding equation (i) and (ii), we get -
\(d^2+(2 b+3 c)^2=0\)