Question:
If \(a\neq 0\) and the line
\[
2bx+3cy+4d=0
\]
passes through the points of intersection of the parabolas
\[
y^2=4ax
\]
and
\[
x^2=4ay,
\]
then
If \(a\neq 0\) and the line
\[
2bx+3cy+4d=0
\]
passes through the points of intersection of the parabolas
\[
y^2=4ax
\]
and
\[
x^2=4ay,
\]
then
Show Hint
When a line passes through the intersection points of two curves, first determine the common points by solving the equations simultaneously. Then substitute those points into the line equation to obtain the required relation among the parameters.
Updated On: Jul 29, 2026
- \[ d^2+(2b-3c)^2=0 \]
- \[ d^2+(3b+2c)^2=0 \]
- \[ d^2+(2b+3c)^2=0 \]
- \[ d^2+(3b-2c)^2=0 \]
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
The required line passes through all common points of the two parabolas. Hence, first find their points of intersection and then impose the condition that both points satisfy the line.
Step 1: Find the points of intersection of the parabolas. Given \[ y^2=4ax \] and \[ x^2=4ay. \] From the first equation, \[ x=\frac{y^2}{4a}. \] Substituting into the second equation, \[ \left(\frac{y^2}{4a}\right)^2 = 4ay. \] \[ \frac{y^4}{16a^2} = 4ay. \] \[ y^4 = 64a^3y. \] \[ y(y^3-64a^3)=0. \] Therefore, \[ y=0 \] or \[ y=4a. \] Hence the corresponding points are \[ (0,0) \] and \[ (4a,4a). \]
Step 2: Use the fact that the line passes through \((0,0)\). Substituting \((0,0)\) into \[ 2bx+3cy+4d=0, \] we get \[ 4d=0. \] \[ d=0. \]
Step 3: Use the point \((4a,4a)\). Substituting \((4a,4a)\), \[ 2b(4a)+3c(4a)+4d=0. \] Since \(d=0\), \[ 8ab+12ac=0. \] \[ 4a(2b+3c)=0. \] Given \[ a\neq 0, \] therefore, \[ 2b+3c=0. \]
Step 4: Combine the two conditions. We have \[ d=0 \] and \[ 2b+3c=0. \] Hence, \[ d^2+(2b+3c)^2=0. \]
Step 5: Write the final answer. \[ \boxed{d^2+(2b+3c)^2=0} \]
Step 1: Find the points of intersection of the parabolas. Given \[ y^2=4ax \] and \[ x^2=4ay. \] From the first equation, \[ x=\frac{y^2}{4a}. \] Substituting into the second equation, \[ \left(\frac{y^2}{4a}\right)^2 = 4ay. \] \[ \frac{y^4}{16a^2} = 4ay. \] \[ y^4 = 64a^3y. \] \[ y(y^3-64a^3)=0. \] Therefore, \[ y=0 \] or \[ y=4a. \] Hence the corresponding points are \[ (0,0) \] and \[ (4a,4a). \]
Step 2: Use the fact that the line passes through \((0,0)\). Substituting \((0,0)\) into \[ 2bx+3cy+4d=0, \] we get \[ 4d=0. \] \[ d=0. \]
Step 3: Use the point \((4a,4a)\). Substituting \((4a,4a)\), \[ 2b(4a)+3c(4a)+4d=0. \] Since \(d=0\), \[ 8ab+12ac=0. \] \[ 4a(2b+3c)=0. \] Given \[ a\neq 0, \] therefore, \[ 2b+3c=0. \]
Step 4: Combine the two conditions. We have \[ d=0 \] and \[ 2b+3c=0. \] Hence, \[ d^2+(2b+3c)^2=0. \]
Step 5: Write the final answer. \[ \boxed{d^2+(2b+3c)^2=0} \]
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