Question:
If \(ax+by=1\) is a normal to the parabola
\[
y^2=4px,
\]
then the condition is
If \(ax+by=1\) is a normal to the parabola
\[ y^2=4px, \] then the condition is
\[ y^2=4px, \] then the condition is
Show Hint
The normal to the parabola \(y^2=4px\) at parameter \(t\) is \(y=-tx+2pt+pt^3\).
Updated On: Jun 15, 2026
- \(4ab=a^2+b^2\)
- \(4pab+ab^3=a^2b^2\)
- \(pa^3=b^2-2pab^2\)
- \(pa^2+4pa=a+b\)
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The Correct Option is C
Solution and Explanation
Step 1: Write the equation of the normal to the parabola.
For the parabola
\[ y^2=4px, \] the normal at parameter \(t\) is
\[ y=-tx+2pt+pt^3 \]
Step 2: Rewrite the given line.
Given line is
\[ ax+by=1 \]
Rearranging,
\[ y=-\frac{a}{b}x+\frac1b \]
Comparing with the normal equation,
\[ t=\frac{a}{b} \]
Also,
\[ \frac1b=2pt+pt^3 \]
Substitute \(t=\dfrac{a}{b}\):
\[ \frac1b = 2p\left(\frac{a}{b}\right) + p\left(\frac{a}{b}\right)^3 \]
\[ \frac1b = \frac{2pa}{b} + \frac{pa^3}{b^3} \]
Step 3: Simplify the equation.
Multiply throughout by \(b^3\):
\[ b^2=2pab^2+pa^3 \]
Rearranging,
\[ pa^3=b^2-2pab^2 \]
Step 4: Final conclusion.
Hence, the required condition is
\[ \boxed{pa^3=b^2-2pab^2} \]
For the parabola
\[ y^2=4px, \] the normal at parameter \(t\) is
\[ y=-tx+2pt+pt^3 \]
Step 2: Rewrite the given line.
Given line is
\[ ax+by=1 \]
Rearranging,
\[ y=-\frac{a}{b}x+\frac1b \]
Comparing with the normal equation,
\[ t=\frac{a}{b} \]
Also,
\[ \frac1b=2pt+pt^3 \]
Substitute \(t=\dfrac{a}{b}\):
\[ \frac1b = 2p\left(\frac{a}{b}\right) + p\left(\frac{a}{b}\right)^3 \]
\[ \frac1b = \frac{2pa}{b} + \frac{pa^3}{b^3} \]
Step 3: Simplify the equation.
Multiply throughout by \(b^3\):
\[ b^2=2pab^2+pa^3 \]
Rearranging,
\[ pa^3=b^2-2pab^2 \]
Step 4: Final conclusion.
Hence, the required condition is
\[ \boxed{pa^3=b^2-2pab^2} \]
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