The line $lx + my + n = 0$ will touch the parabola $y^2 = 4ax$ if $ln = am^2$.

The line lx + my + n = 0 will touch the parabola y² = 4ax if ln = am².

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lx + my + n = 0

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Concept understanding — Parabola Standard Form

Standard Equations of a Parabola

A parabola is the set of all points in a plane that are equidistant from a fixed point (the focus) and a fixed line (the directrix). To turn this definition into a clean equation, NCERT places the parabola in the simplest position: vertex at the origin with its axis along a coordinate axis. The equations you get are called the standard equations.

The four standard forms

Depending on which way the parabola opens, there are four standard equations. In each, $a > 0$ .

y^2 = 4ax

Equation	Opens	Focus	Directrix
$y^2 = 4ax$	right	$(a, 0)$	$x = -a$
$y^2 = -4ax$	left	$(-a, 0)$	$x = a$
$x^2 = 4ay$	up	$(0, a)$	$y = -a$
$x^2 = -4ay$	down	$(0, -a)$	$y = a$

For all four the vertex is at the origin $(0,0)$ and the axis of the parabola is a coordinate axis.

Where $y^2 = 4ax$ comes from

Take the focus at $F(a, 0)$ and the directrix as the line $x = -a$ . For a point $P(x, y)$ on the parabola, its distance to the focus equals its distance to the directrix:

\sqrt{(x - a)^2 + y^2} = x + a.

Squaring both sides:

(x - a)^2 + y^2 = (x + a)^2,

and expanding gives $y^2 = 4ax$ . The other three forms follow by turning the focus in a different direction.

Latus rectum

The latus rectum is the chord through the focus, perpendicular to the axis, with both ends on the parabola. For every standard parabola its length is $4a$ — the very same $4a$ that appears in the equation, which makes it quick to read off.

Worked example

For the parabola $y^2 = 12x$ , compare with $y^2 = 4ax$ : here $4a = 12$ , so $a = 3$ .

Vertex: $(0, 0)$
Focus: $(a, 0) = (3, 0)$
Directrix: $x = -3$ …

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A line touches a parabola when it meets the curve at exactly one point. Substituting the line $lx+my+n=0$ into $y^2=4ax$ gives a quadratic in $y$ ; forcing its discriminant to zero yields the tangency condition $ln = am^2$ , so the statement is TRUE.

The claim is that the line $lx + my + n = 0$ is tangent to the parabola $y^2 = 4ax$ exactly when $ln = am^2$ . "Touches" means tangent — the line and the curve share exactly one common point. Algebraically, if we solve the two equations together we get a quadratic, and "exactly one solution" means its discriminant is zero. Let us derive the condition and check it against the statement.

Step 1 — Set up the intersection

We want the points common to the line and the parabola. From the line $lx + my + n = 0$ , express $x$ in terms of $y$ (taking $l \neq 0$ , the case of a genuine slanted/vertical tangent):

x = -\frac{my + n}{l}.

Step 2 — Substitute into the parabola

Put this $x$ into $y^2 = 4ax$ :

y^2 = 4a\left(-\frac{my + n}{l}\right) = -\frac{4a(my+n)}{l}.

Multiply through by $l$ and collect all terms on one side:

ly^2 + 4amy + 4an = 0.

This is a quadratic in $y$ , of the form $Ay^2 + By + C = 0$ with

A = l, \qquad B = 4am, \qquad C = 4an.

Step 3 — Impose tangency (discriminant $= 0$ )

The line touches the parabola when this quadratic has a repeated root, i.e. its discriminant vanishes:

\Delta = B^2 - 4AC = (4am)^2 - 4(l)(4an) = 0.

Expand:

16a^2m^2 - 16aln = 0.

Step 4 — Simplify

Factor out $16a$ :

$16a\left(am^2 - ln\right) = 0.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
$P(x, y, z)$
›Reveal solutionSolution
The perpendicular distance of $P$ from the X-axis is $\sqrt{y^2+z^2}$ and from the YZ-plane is $|x|$ . Setting their ratio to $2:3$ gives $3\sqrt{y^2+z^2}=2|x|$ , and squaring yields $9(y^2+z^2)=4x^2$ , i.e. $4x^2-9y^2-9z^2=0$ - option (A).

Concept & Intuition

We need the locus of $P(x,y,z)$ for which
$\frac{\text{distance of }P\text{ from the X-axis}}{\text{distance of }P\text{ from the YZ-plane}}=\frac{2}{3}.$

The X-axis is the set of points $(t,0,0)$ , so the perpendicular distance from $P$ to it is the distance measured in the $y$ - $z$ directions: $d_X=\sqrt{y^2+z^2}$ .

The YZ-plane is $x=0$ , so the perpendicular distance from $P$ to it is $d_{YZ}=|x|$ .

Step-by-Step Derivation

Write the two distances.

$d_X=\sqrt{y^2+z^2}$ and $d_{YZ}=|x|$ .

Impose the ratio.

$\frac{\sqrt{y^2+z^2}}{|x|}=\frac{2}{3}\ \Longrightarrow\ 3\sqrt{y^2+z^2}=2|x|.$

Square both sides (both are non-negative, so no extraneous roots):

$9(y^2+z^2)=4x^2.$

Rearrange to standard form. $4x^2-9y^2-9z^2=0.$ …
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TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
$y = x^3 - 9x^2 + cx - d$
›Reveal solutionSolution

Killing the $X^2$ term forces $h=3$ and killing the $X$ term forces $c=27$ ; the official key marks the value $13$ , option (B).

Shift the origin. Put $x=X+h,\ y=Y+5$ into $y=x^3-9x^2+cx-d$ and require the result to be $Y=X^3$ .

Vanishing $X^2$ term. The coefficient of $X^2$ is $3h-9=0$ , so
$h=3.$
Vanishing $X$ term. The coefficient of $X$ is $3h^2-18h+c=27-54+c=0$ , so
$c=27.$
Equivalently $x^3-9x^2+27x-d=(x-3)^3+(27-d)$ , confirming the depressed-cubic shift is exact. …
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TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$\theta$
›Reveal solutionSolution
The key is to use the focus–directrix property of a hyperbola: for any point P on the hyperbola, SP = e·(distance from P to the corresponding directrix). Combining this with the given SQ distance and the coordinates of Q yields the eccentricity and then the parameter θ.

Concept & Intuition

We have a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{9} = 1$ . Its foci are at $(\pm ae, 0)$ where $e = \sqrt{1 + \frac{b^2}{a^2}}$ and here $b^2 = 9$ . The focus on the positive X‑axis is $S = (ae, 0)$ .

We are given a point $P(\theta)$ on the hyperbola — this notation usually means the parametric form: $P = (a\sec\theta, 3\tan\theta)$ .

We know $SP = 6$ and $SQ = \sqrt{26}$ with $Q = (0,1)$ . The distance $SQ$ will let us find $ae$ (the x‑coordinate of S). Then using $SP = 6$ and the parametric coordinates, we can solve for $\theta$ .

Step‑by‑step solution

Find the focus coordinate $S = (ae, 0)$ using $SQ = \sqrt{26}$ . $Q = (0,1)$ , so

$SQ^2 = (ae - 0)^2 + (0 - 1)^2 = a^2 e^2 + 1 = 26.$
Hence
$a^2 e^2 = 25 \quad\Rightarrow\quad ae = 5.$
So $S = (5, 0)$ .

Relate $a$ and $e$ . For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ , we have $e = \sqrt{1 + \frac{b^2}{a^2}}$ . Here $b^2 = 9$ , so

$e = \sqrt{1 + \frac{9}{a^2}}.$
But we also have $ae = 5$ . Substituting:
$a \sqrt{1 + \frac{9}{a^2}} = 5 \quad\Rightarrow\quad \sqrt{a^2 + 9} = 5.$
Squaring: $a^2 + 9 = 25 \Rightarrow a^2 = 16 \Rightarrow a = 4$ (positive).

Then $e = \frac{5}{a} = \frac{5}{4}$ .

Parametric form of P. The hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ has parametric coordinates

$P = (4\sec\theta,\; 3\tan\theta).$

Use $SP = 6$ .

$SP^2 = (4\sec\theta - 5)^2 + (3\tan\theta - 0)^2 = 36.$
Expand:
$16\sec^2\theta - 40\sec\theta + 25 + 9\tan^2\theta = 36.$
Recall $\tan^2\theta = \sec^2\theta - 1$ . Substitute:
$16\sec^2\theta - 40\sec\theta + 25 + 9(\sec^2\theta - 1) = 36.$
Simplify:
$25\sec^2\theta - 40\sec\theta + 16 = 36.$
So
$25\sec^2\theta - 40\sec\theta - 20 = 0.$
Divide by 5:
$5\sec^2\theta - 8\sec\theta - 4 = 0.$

Solve the quadratic in $\sec\theta$ . $\sec\theta = \frac{8 \pm \sqrt{64 + 80}}{10} = \frac{8 \pm \sqrt{144}}{10} = \frac{8 \pm 12}{10}.$ …
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TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
$x \cos \alpha + y \sin \alpha = 2\sqrt{3}$
›Reveal solutionSolution

Impose the ellipse tangency condition $c^2=a^2m^2+b^2$ on the given normal-form line; it reduces to $\sin^2\alpha=\tfrac{1}{2}$ , so the acute angle is $\alpha=\tfrac{\pi}{4}$ .

Step 1 — line in slope form. $x\cos\alpha+y\sin\alpha=2\sqrt{3}$ gives
$y=-\cot\alpha\,x+\frac{2\sqrt{3}}{\sin\alpha},\qquad m=-\cot\alpha,\quad c=\frac{2\sqrt{3}}{\sin\alpha}.$
Step 2 — tangency to $\dfrac{x^2}{16}+\dfrac{y^2}{8}=1$ . Here $a^2=16,\,b^2=8$ , and tangency requires $c^2=a^2m^2+b^2$ :
$\frac{12}{\sin^2\alpha}=16\cot^2\alpha+8.$
Step 3 — solve. Using $\cot^2\alpha=\dfrac{1-\sin^2\alpha}{\sin^2\alpha}$ and multiplying by $\sin^2\alpha$ : …
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Q.The line lx+my+n=0lx + my + n = 0lx+my+n=0 will touch the parabola y2=4axy^2 = 4axy2=4ax if ln=am2ln = am^2ln=am2.

Step 1 — Set up the intersection

Step 2 — Substitute into the parabola

Step 3 — Impose tangency (discriminant =0= 0=0)

Step 4 — Simplify

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Q.The line $lx + my + n = 0$ will touch the parabola $y^2 = 4ax$ if $ln = am^2$ .

Step 3 — Impose tangency (discriminant $= 0$ )

Unlock everything free for 14 days