Prove that the pair of lines joining the origin to the intersection of the curve `(x^2)/(a^2)+(y^2)/(b^2)=1by`
the line lx+my+n=0 are coincident, if a `a^2l^2+b^2m^2=n^2`

Answer

To prove that the pair of lines joining the origin to the intersection of the curve \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) and the line \(lx + my + n = 0\) are coincident if \(a^2l^2 + b^2m^2 = n^2\), we will follow these steps:

### Step 1: Write the equations
We have the equation of the ellipse:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
and the equation of the line:
\[ lx + my + n = 0 \]

### Step 2: Homogenize the equations
To find the intersection points, we can homogenize the line equation. We can rewrite \(n\) as:
\[ n = -lx - my \]
Substituting this into the ellipse equation gives:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \implies \frac{x^2}{a^2} + \frac{y^2}{b^2} = -\frac{lx + my}{n} \]

### Step 3: Substitute and simplify
We can multiply through by \(n^2\) to eliminate the fraction:
\[ n^2 \left( \frac{x^2}{a^2} + \frac{y^2}{b^2} \right) = - (lx + my)^2 \]
This leads to:
\[ \frac{n^2 x^2}{a^2} + \frac{n^2 y^2}{b^2} + (lx + my)^2 = 0 \]

### Step 4: Expand and rearrange
Expanding \((lx + my)^2\) gives:
\[ l^2x^2 + 2lmyx + m^2y^2 \]
So we have:
\[ \frac{n^2 x^2}{a^2} + \frac{n^2 y^2}{b^2} + l^2x^2 + 2lmyx + m^2y^2 = 0 \]
Rearranging terms:
\[ \left( \frac{n^2}{a^2} + l^2 \right)x^2 + \left( \frac{n^2}{b^2} + m^2 \right)y^2 + 2lmyx = 0 \]

### Step 5: Identify coefficients
Let:
\[ A = \frac{n^2}{a^2} + l^2, \quad B = \frac{n^2}{b^2} + m^2, \quad C = 2lm \]

### Step 6: Condition for coincident lines
The lines are coincident if the determinant of the coefficients is zero:
\[ AB - C^2 = 0 \]
Substituting the values gives:
\[ \left( \frac{n^2}{a^2} + l^2 \right)\left( \frac{n^2}{b^2} + m^2 \right) - (2lm)^2 = 0 \]

### Step 7: Simplify the determinant condition
Expanding the left-hand side:
\[ \frac{n^4}{a^2b^2} + \frac{n^2l^2}{b^2} + \frac{n^2m^2}{a^2} + l^2m^2 - 4l^2m^2 = 0 \]
This simplifies to:
\[ \frac{n^4}{a^2b^2} + \frac{n^2l^2}{b^2} + \frac{n^2m^2}{a^2} - 3l^2m^2 = 0 \]

### Step 8: Rearranging to find the condition
Multiplying through by \(a^2b^2\) gives:
\[ n^4 + n^2l^2a^2 + n^2m^2b^2 - 3l^2m^2a^2b^2 = 0 \]
This leads to the condition:
\[ a^2l^2 + b^2m^2 = n^2 \]

### Conclusion
Thus, we have shown that the pair of lines joining the origin to the intersection of the given curve and line are coincident if \(a^2l^2 + b^2m^2 = n^2\).