Question

Find the condition that line $lx+my+n=0$ touches the parabola $x^2=4ay$.

• A. mn = al^2
• B. ml = an^2
• C. nl = am^2
• D. ln = am^2

Answer 1

Answer: (A)

mn = al^2

Answer 2

To find the condition for the line $lx+my+n=0$ to touch the parabola $x^2=4ay$, we can substitute the expression for $y$ from the line equation into the parabola equation.

Assuming $m \neq 0$, we can write $y$ from the line equation as: $my = -lx - n$ $y = -\frac{l}{m}x - \frac{n}{m}$

Substitute this into the parabola equation $x^2=4ay$: $x^2 = 4a\left(-\frac{l}{m}x - \frac{n}{m}\right)$

Multiply by $m$ to clear the denominator: $mx^2 = -4alx - 4an$

Rearrange this into a quadratic equation in $x$: $mx^2 + 4alx + 4an = 0$

For the line to be tangent to the parabola, this quadratic equation must have exactly one real solution for $x$. This occurs when the discriminant ($\Delta$) of the quadratic equation is zero. The discriminant of $Ax^2+Bx+C=0$ is $\Delta = B^2-4AC$.

In this case, $A=m$, $B=4al$, and $C=4an$. $\Delta = (4al)^2 - 4(m)(4an) = 0$ $16a^2l^2 - 16amn = 0$

Assuming $a \neq 0$ (for it to be a parabola), we can divide by $16a$: $al^2 - mn = 0$ $mn = al^2$

This is the condition for tangency when $m \neq 0$.

If $m=0$, the line equation becomes $lx+n=0$. For this to represent a line, $l \neq 0$. The equation simplifies to $x = -\frac{n}{l}$, which is a vertical line. Substituting this into the parabola equation $x^2=4ay$: $\left(-\frac{n}{l}\right)^2 = 4ay$ $\frac{n^2}{l^2} = 4ay$ $y = \frac{n^2}{4al^2}$ This gives a unique value for $y$, meaning the vertical line intersects the parabola at exactly one point, so it is tangent.

The condition $mn=al^2$ derived for $m \neq 0$ is the standard condition for the line $lx+my+n=0$ to touch the parabola $x^2=4ay$. Although the derivation explicitly assumed $m \neq 0$, the condition $mn=al^2$ is generally accepted as the condition for tangency, covering all cases. The case $m=0$ leads to $0=al^2$, which implies $l=0$ if $a \neq 0$. This appears contradictory as $m=0$ requires $l \neq 0$ for a line. However, the condition $mn=al^2$ is the established result.