Question

If the lines represented by
\[ ax^2+2hxy+by^2+2gx+2fy+c=0 \] intersect on the \(x\)-axis, which of the following is in general incorrect?

• A. \(abc=2fgh\)
• B. \(g^2=ac\)
• C. \(af^2=ch^2\)
• D. \(af^2+ch^2=2fgh\)

Hint:

For a pair of straight lines, the point of intersection is found using \(ax+hy+g=0\) and \(hx+by+f=0\). If the intersection lies on the \(x\)-axis, put \(y=0\).

Answer 1

The Correct Option is A

Step 1: Use the condition that the point of intersection lies on the \(x\)-axis.
Let the point of intersection be \((\alpha,0)\).
For the pair of lines, the point of intersection satisfies
\[ ax+hy+g=0 \] and
\[ hx+by+f=0 \]
Putting \(x=\alpha,\;y=0\), we get
\[ a\alpha+g=0 \] and
\[ h\alpha+f=0 \]
So,
\[ \alpha=-\frac{g}{a} \] and
\[ \alpha=-\frac{f}{h} \]
Hence,
\[ \frac{g}{a}=\frac{f}{h} \]
\[ af=gh \]

Step 2: Use the fact that \((\alpha,0)\) lies on the given equation.
Substitute \((\alpha,0)\) in the given equation:
\[ a\alpha^2+2g\alpha+c=0 \]
Using \(\alpha=-\dfrac{g}{a}\),
\[ a\left(\frac{g^2}{a^2}\right)-\frac{2g^2}{a}+c=0 \]
\[ \frac{g^2}{a}-\frac{2g^2}{a}+c=0 \]
\[ -\frac{g^2}{a}+c=0 \]
\[ g^2=ac \]
So, option (2) is correct.

Step 3: Derive another relation.
From
\[ af=gh \] we get
\[ f=\frac{gh}{a} \]
Now,
\[ af^2=a\left(\frac{g^2h^2}{a^2}\right) \]
\[ =\frac{g^2h^2}{a} \]
Using \(g^2=ac\),
\[ af^2=\frac{ach^2}{a} \]
\[ af^2=ch^2 \]
So, option (3) is correct.

Step 4: Check option (4).
Since
\[ af^2=ch^2 \] we have
\[ af^2+ch^2=2ch^2 \]
Also, from \(f=\dfrac{gh}{a}\),
\[ 2fgh=2\cdot\frac{gh}{a}\cdot g\cdot h \]
\[ =\frac{2g^2h^2}{a} \]
Using \(g^2=ac\),
\[ 2fgh=\frac{2ach^2}{a}=2ch^2 \]
Thus,
\[ af^2+ch^2=2fgh \]
So, option (4) is correct.

Step 5: Identify the incorrect statement.
The relation
\[ abc=2fgh \] does not follow in general from the given condition.
Hence, the generally incorrect statement is
\[ \boxed{abc=2fgh} \]