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

The correct option is C

$d^2+{(2b+3c)}^2=0$

Explantion for the correct option:

Step-1: Finding solution of the parabola:

Given equation of parabola are $x^2=4ay,$ & $y^2=4ax$

And equation of line $2bx+3cy+4d=0...\left(i\right)$

On solving both equation

$$\begin{gathered} x^2=4ay...\left(i\right) \\ {y}^2=4ax...\left(ii\right) \\ \Rightarrow x=\frac{y^2}{4a} \\ \mathrm{Put}\mathrm{in}\left(\mathrm{i}\right) \\ \frac{y^4}{16a^2}=4ay \\ \Rightarrow y^4=64a^{3}y \\ \Rightarrow y(y^3-64a^3)=0 \\ \Rightarrow y=0ory=4a \end{gathered}$$

Therefore points are $(0,0)\&(4a,4a)$

Step 2. Solving equation of the line:

Put the above point of $x,y$ in equation $\left(i\right)$

$$\begin{gathered} 2bx+3cy+4d=0 \\ Putx=0\&y=0 \\ \Rightarrow d=0 \\ Alsox=4a,y=4a\&d=0 \\ 2ab+3ac=0 \\ \Rightarrow 2b+3c=0 \\ \mathrm{Therefore},d^2+{(2b+3c)}^2=0 \end{gathered}$$

Hence, correct option is $\mathbf{\left(}\mathit{C}\mathbf{\right)}$.