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

Showing the 12 most recent of 48 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $(\alpha, \beta)$ is the mid point of a focal chord of the parabola $x^2 = 20y$ whose slope is 1, then $3\alpha =$ (A) $\beta + 2$ (B) $4\beta$ (C) $\beta + 4$ (D) $2\beta$

   ›Reveal solutionSolution

   Parametrizing the parabola and using the focal-chord property $t_1t_2=-1$ gives the midpoint $(10,15)$, so $3\alpha=2\beta$.

   Concept and Intuition

   For $x^2=4ay$, points can be written as $(2at,at^2)$. A chord through the focus $(0,a)$ has parameters satisfying $t_1t_2=-1$ — a direct analogue of the more familiar $y^2=4ax$ focal-chord property. The slope of the chord equals $(t_1+t_2)/2$.

   Step-by-Step Solution

   1. $x^2=20y \Rightarrow 4a=20\Rightarrow a=5$; points parametrized as $(10t,5t^2)$.
   2. Focal chord condition: $t_1t_2=-1$.
   3. Chord slope $=\dfrac{5t_2^2-5t_1^2}{10t_2-10t_1}=\dfrac{t_1+t_2}{2}=1\Rightarrow t_1+t_2=2$.
   4. Midpoint $x$-coordinate: $\alpha=\dfrac{10t_1+10t_2}{2}=5(t_1+t_2)=10$.
   5. $t_1^2+t_2^2=(t_1+t_2)^2-2t_1t_2=4-2(-1)=6$. …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.A chord of the parabola $y = x^2 + 3x + 2$, among the following is (A) $8x - 4y + 7 = 0$ (B) $x + y + 2 = 0$ (C) $x - y + 1 = 0$ (D) $x - y + 2 = 0$

   ›Reveal solutionSolution

   A chord meets the curve in two distinct points; substituting each option's line into the parabola shows (A),(B),(C) are tangents (repeated root) while (D) meets it at $x=0$ and $x=-2$ — a genuine chord.

   Concept and Intuition

   For a line to be a chord of $y=x^2+3x+2$ it must intersect the parabola at two distinct real points. Substituting $y$ from the line into the parabola's equation gives a quadratic in $x$; a positive discriminant (two real roots) means a chord, zero discriminant means the line is tangent, and a negative discriminant means no intersection at all.

   Step-by-Step Solution

   1. (A) $8x-4y+7=0 \Rightarrow y=2x+\tfrac74$. Set equal: $x^2+3x+2 = 2x+\tfrac74 \Rightarrow x^2+x+\tfrac14=0$, discriminant $=1-1=0$ — tangent.
   2. (B) $x+y+2=0 \Rightarrow y=-x-2$. Set equal: $x^2+3x+2=-x-2 \Rightarrow x^2+4x+4=0=(x+2)^2$ — tangent.
   3. (C) $x-y+1=0 \Rightarrow y=x+1$. Set equal: $x^2+3x+2=x+1 \Rightarrow x^2+2x+1=0=(x+1)^2$ — tangent. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $y = mx + c, m > 0$ is a common tangent to the parabolas $y^2 = 8x$ and $y^2 = 1 + 4x$, then $m + c =$ (A) $\dfrac{\sqrt{5}}{2} + \dfrac{9}{2\sqrt{5}}$ (B) $5$ (C) $\dfrac{\sqrt{5}}{8} + \dfrac{2}{\sqrt{5}}$ (D) $3$

   ›Reveal solutionSolution

   Writing the standard tangent condition $c=a/m$ for each (shifted) parabola and equating the two expressions for $c$ gives $m=2,\,c=1$, so $m+c=3$.

   Concept and Intuition

   For $y^2=4ax$, the line $y=mx+c$ is tangent exactly when $c=\dfrac am$. When a parabola is shifted, $y^2=4a(x-h)$, the same condition applies to $x-h$, giving $c=am+\ldots$ — more precisely $y=mx+c$ is tangent to $y^2=4a(x-h)$ iff $c=\dfrac{a}{m}+mh$. A common tangent to two parabolas must satisfy both tangency conditions simultaneously, giving one equation in $m$.

   Step-by-Step Solution

   1. $y^2=8x$ is $y^2=4(2)x$ with $a=2$; tangent $y=mx+c$ needs $c=\dfrac{2}{m}$.
   2. $y^2=1+4x = 4\left(x+\dfrac14\right)$ is $y^2=4a X$ with $a=1$, $X=x+\frac14$; tangent condition $c'=a/m$ in terms of $X$ becomes, back in $x$: $y=mx+\left(\dfrac{m}{4}+\dfrac1m\right)$, i.e. $c=\dfrac m4+\dfrac1m$. …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Through the focus of the parabola $x^2 - 4x - 8y + 44 = 0$, if tangents are drawn to another parabola $y^2 = 20x$, then the sum of the $Y$-coordinates of the points of contact of these tangents is (A) $20$ (B) $8$ (C) $14$ (D) $32$

   ›Reveal solutionSolution

   The focus of the first parabola is $(2,7)$; solving for the tangent parameters $t$ from this external point to $y^2=20x$ gives contact points with $Y$-coordinates summing to $14$.

   Concept and Intuition

   For $y^2=4ax$, points on the curve are $(at^2,2at)$ and the tangent there is $ty=x+at^2$. From an external point, substituting its coordinates into the tangent equation gives a quadratic in $t$ whose two roots are the parameters of the two points of contact — Vieta's formulas then give sums/products of those parameters without solving explicitly (though direct solving works fine here too).

   Step-by-Step Solution

   1. Complete the square: $x^2-4x-8y+44=0 \Rightarrow (x-2)^2-4-8y+44=0 \Rightarrow (x-2)^2=8y-40=8(y-5)$.
   2. This is $X^2=4aY$ with $X=x-2,\,Y=y-5,\,4a=8\Rightarrow a=2$. Vertex $(2,5)$, axis vertical, opens upward; focus at $(2,5+2)=(2,7)$.
   3. For $y^2=20x=4(5)x$, points are $(5t^2,10t)$ with tangent $ty=x+5t^2$.
   4. This tangent passes through the focus $(2,7)$: $7t=2+5t^2 \Rightarrow 5t^2-7t+2=0$. …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If the length of the tangent at a point on the parabola $y^2 = 4ax$ is $4a\sqrt{5}$, then the length of the sub normal at that point is (A) $4a$ (B) $a$ (C) $8a$ (D) $2a$

   ›Reveal solutionSolution

   The subnormal of the parabola $y^2=4ax$ is the constant $2a$ at every point on the

   curve — a standard property — so the tangent-length data, while consistent with a real

   point ($x=4a$), isn't even needed to get the subnormal.

   Concept and Intuition

   For a curve $y=f(x)$ at $(x_1,y_1)$ with slope $m=\frac{dy}{dx}$: the subnormal is the

   projection of the normal's segment onto the x-axis, of length $|m\,y_1|$. For

   $y^2=4ax$, implicit differentiation gives $2y\,y'=4a \Rightarrow y'=\dfrac{2a}{y_1}$, so

   the subnormal is $y_1\cdot\dfrac{2a}{y_1}=2a$ — a constant, true at every point of the

   parabola, a signature feature of this curve.

   Step-by-Step Solution

   1. Differentiate $y^2=4ax$: $2y\,\dfrac{dy}{dx}=4a \Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{y}$.
   2. Subnormal $=y\cdot\dfrac{dy}{dx}=y\cdot\dfrac{2a}{y}=2a$ — independent of $y$.
   3. (Consistency check) Length of tangent at $(x_1,y_1)$ works out to $2\sqrt{x_1(x_1+a)}$; setting this to $4a\sqrt5$ gives $x_1(x_1+a)=20a^2$, i.e. …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.A chord is drawn through the focus of the parabola $y^2 = 6x$ such that its perpendicular distance from the vertex is $\dfrac{\sqrt{5}}{2}$. Then its slope can be (A) $\dfrac{2}{\sqrt{3}}$ (B) $\dfrac{2}{\sqrt{5}}$ (C) $\dfrac{\sqrt{3}}{2}$ (D) $\dfrac{\sqrt{5}}{2}$

   ›Reveal solutionSolution

   Setting up the focal-chord line through $\left(\tfrac32,0\right)$ and equating its perpendicular distance from the origin to $\tfrac{\sqrt5}{2}$ gives $m=\pm\tfrac{\sqrt5}{2}$. Answer: (D).

   Concept and Intuition

   Every chord through the focus can be described by its slope $m$; its perpendicular distance from any fixed point (here, the vertex) can be computed with the standard point-to-line distance formula. This converts a geometric condition (given perpendicular distance) directly into an equation for $m$.

   Step-by-Step Solution

   1. Parabola $y^2=6x$ compared with $y^2=4ax$ gives $4a=6\Rightarrow a=\dfrac32$. Focus $=\left(\dfrac32,0\right)$.
   2. A chord through the focus with slope $m$: $y=m\left(x-\dfrac32\right)\Rightarrow mx-y-\dfrac{3m}{2}=0$.
   3. Perpendicular distance from vertex $(0,0)$:

   $\frac{\left|0-0-\frac{3m}{2}\right|}{\sqrt{m^2+1}}=\frac{\frac{3|m|}{2}}{\sqrt{m^2+1}}$

   4. Set equal to $\dfrac{\sqrt5}{2}$: …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.A perpendicular is drawn through the vertex 'O' of the parabola $y^2 = 8x$ to any non-vertical tangent meeting it at $P$ and the parabola at $Q$. Then $OP \cdot OQ =$ (A) $16$ (B) $12$ (C) $6$ (D) $24$

   ›Reveal solutionSolution

   Using the tangent-line and foot-of-perpendicular formulas, then intersecting the perpendicular line with the parabola again, gives a constant product $OP\cdot OQ=16$ independent of $m$. Answer: (A).

   Concept and Intuition

   Any non-vertical tangent to $y^2=4ax$ can be written as $y=mx+\dfrac{a}{m}$. Dropping a perpendicular from the vertex to this tangent gives a foot $P$; extending that same perpendicular line hits the parabola again at a second point $Q$. Remarkably, the product $OP\cdot OQ$ turns out to be independent of the tangent's slope — a nice invariant of the parabola.

   Step-by-Step Solution

   1. Parabola $y^2=8x \Rightarrow 4a=8\Rightarrow a=2$. Non-vertical tangent: $y=mx+\dfrac2m$, i.e., $mx-y+\dfrac2m=0$.
   2. Foot of perpendicular from origin to line $ax+by+c=0$ is $\left(\dfrac{-ac}{a^2+b^2},\dfrac{-bc}{a^2+b^2}\right)$. Here the line coefficients are $(m,-1,\tfrac2m)$:

   $P=\left(\frac{-2}{m^2+1},\ \frac{2}{m(m^2+1)}\right)$

   3. $OP^2=\frac{4}{(m^2+1)^2}+\frac{4}{m^2(m^2+1)^2}=\frac{4(m^2+1)}{m^2(m^2+1)^2}=\frac{4}{m^2(m^2+1)}$ so $OP=\dfrac{2}{|m|\sqrt{m^2+1}}$. …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.PSQ is a focal chord of the parabola $y^2=12x$. A, B are respectively the feet of the perpendicular S drawn from P and Q on to the directrix of the parabola. If the length of AB is $7\sqrt{3}$ and $P=(3t^2,6t)$ $(0<t<1)$, then $t=$ (A) $\dfrac{2}{\sqrt{3}}$ (B) $\dfrac{\sqrt{3}}{2}$ (C) $\dfrac{\sqrt{2}}{3}$ (D) $\dfrac{\sqrt{5}}{4}$

   ›Reveal solutionSolution

   Use the focal-chord parameter relation $t_1t_2=-1$ and the fact that the feet on the directrix have the same $y$-coordinates as $P,Q$; solving the resulting quadratic gives $t=\sqrt3/2$.

   Concept and Intuition

   For the parabola $y^2=4ax$ parametrized as $(at^2,2at)$, if $P,Q$ are the ends of a focal chord with parameters $t_1,t_2$, then $t_1t_2=-1$ (a standard focal-chord property). Since the directrix is the vertical line $x=-a$, the feet of the perpendiculars from any point onto the directrix simply keep that point's $y$-coordinate — so the distance $AB$ between the two feet equals $|y_P-y_Q|$.

   Step-by-Step Solution

   1. $y^2=12x=4(3)x\Rightarrow a=3$. So $P=(3t^2,6t)$ matches the standard parametrization with this $a$.
   2. Since $PSQ$ is a focal chord, $Q$'s parameter is $t'=-1/t$, so $Q=\left(\dfrac{3}{t^2},-\dfrac{6}{t}\right)$.
   3. $A=(-3,6t)$, $B=\left(-3,-\dfrac6t\right)$ (feet on directrix $x=-3$, same $y$ as $P,Q$).
   4. $AB=\left|6t-\left(-\dfrac6t\right)\right|=6\left(t+\dfrac1t\right)$ (positive since $0<t<1$).
   5. Set $6\left(t+\dfrac1t\right)=7\sqrt3\Rightarrow t+\dfrac1t=\dfrac{7\sqrt3}{6}$. …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Let O, S be the vertex and focus of the parabola $y^2=4ax$ respectively and $x=k$ be its double ordinate of length $2\sqrt{6}a$. If the line $x=k$ cuts the X-axis at P, then length of the double ordinate drawn through O to the parabola having P and S as vertex and focus is (A) $4\sqrt{6}a$ (B) $4\sqrt{3}a$ (C) $2\sqrt{2}a$ (D) $2\sqrt{3}a$

   ›Reveal solutionSolution

   Locate $P$ from the given double-ordinate length, build the new (leftward-opening) parabola with vertex $P$ and focus $S$, then find its double ordinate through $O$: $2\sqrt3a$.

   Concept and Intuition

   For $y^2=4ax$, the double ordinate at $x=x_0$ has length $2\sqrt{4ax_0}=4\sqrt{ax_0}$ (twice the $y$-value there). A new parabola is fully determined by its vertex and focus: if the focus lies to the left of the vertex, the parabola opens leftward, with semi-latus-rectum parameter $a'=$ distance between vertex and focus, giving equation $y^2=-4a'(x-x_{\text{vertex}})$.

   Step-by-Step Solution

   1. Double ordinate at $x=k$ has length $4\sqrt{ak}=2\sqrt6\,a\Rightarrow\sqrt{ak}=\dfrac{\sqrt6}{2}a\Rightarrow ak=\dfrac32a^2\Rightarrow k=\dfrac32a$.
   2. $P=(k,0)=\left(\dfrac32a,0\right)$ (where $x=k$ meets the X-axis).
   3. Original parabola's focus: $S=(a,0)$.
   4. New parabola has vertex $P=\left(\dfrac32a,0\right)$ and focus $S=(a,0)$; since $S$ is to the left of $P$, it opens in the $-x$ direction with $a'=PS=\dfrac32a-a=\dfrac{a}{2}$. …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
    Q.The equation of the pair of lines joining the vertex of the parabola $y^2=12x$ and the points of intersection of this parabola and the focal chord drawn to this parabola having slope 2 is (A) $4x^2-2xy-y^2=0$ (B) $4x^2+2xy+y^2=0$ (C) $x^2-2xy+4y^2=0$ (D) $x^2+2xy-4y^2=0$

    ›Reveal solutionSolution

    Homogenise the parabola's equation using the focal-chord line (made $=1$) to get the combined equation of the two lines from the vertex to the chord's endpoints: $4x^2-2xy-y^2=0$.

    Concept and Intuition

    To get the pair of lines joining the origin (vertex) to the two points where a curve meets a line, we make the line's equation equal to $1$ and multiply it into the lower-degree terms of the curve's equation so that every term becomes degree 2 — this is the standard "homogenising" trick, valid because at the intersection points the line equation genuinely equals $1$, so multiplying by it doesn't change the value there but raises the degree everywhere.

    Step-by-Step Solution

    1. Parabola $y^2=12x$ ($4a=12\Rightarrow a=3$), vertex at origin, focus at $(a,0)=(3,0)$.
    2. Focal chord with slope $2$ through the focus: $y-0=2(x-3)\Rightarrow2x-y-6=0\Rightarrow2x-y=6\Rightarrow\dfrac{2x-y}{6}=1$.
    3. Homogenise $y^2=12x$: replace the "$12x$" (degree 1) by $12x\times\left(\dfrac{2x-y}{6}\right)$ (multiplying by the line expression, which equals 1 exactly at the intersection points), making the whole equation degree 2: …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
    Q.If three normals are drawn to the parabola $y^2=x$ from a point $(C,0)$, then (A) $C>\dfrac{1}{2}$ (B) $C<\dfrac{1}{2}$ (C) $C=\dfrac{1}{2}$ (D) $C=\dfrac{1}{4}$

    ›Reveal solutionSolution

    Tests the condition for three real normals to a parabola from a point on its axis; the answer is $C>\tfrac12$.

    Concept and Intuition

    From a point on the axis of a parabola, either one or three real normals can be drawn depending on how far the point is from the vertex — this is exactly analogous to how many perpendiculars from an axial point meet the curve. The threshold distance is $2a$ (where $4a$ is the coefficient of $x$ in $y^2=4ax$).

    Step-by-Step Solution

    1. Write $y^2=x$ as $y^2=4ax$ with $4a=1\Rightarrow a=\dfrac14$.
    2. The normal to $y^2=4ax$ at the parameter point $(at^2,2at)$ is $y=-tx+2at+at^3$.
    3. For this normal to pass through $(C,0)$: $0=-tC+2at+at^3=t(at^2+2a-C)$.
    4. One root is always $t=0$ (the normal at the vertex). The other two roots satisfy $at^2=C-2a$, i.e. $t^2=\dfrac{C-2a}{a}$. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
    Q.If $y = mx + \frac{3}{m}$ is a tangent drawn to the parabola $y^2 = 4ax$ at the point $P(3, \beta)$ ($\beta < 0$), then $3m - \beta =$ (A) $2a$ (B) $ma$ (C) $a$ (D) $\frac{|\beta|}{a}$

    ›Reveal solutionSolution

    This tests the slope-form tangent to a parabola. Matching the given tangent to the standard form pins down $a$, then the point $P$ pins down $m$ and $\beta$, giving $3m-\beta=a$.

    Concept and Intuition

    Every tangent of slope $m$ to $y^2=4ax$ can be written as $y=mx+\frac{a}{m}$, and it touches the parabola exactly at $\left(\frac{a}{m^2},\frac{2a}{m}\right)$. Recognising the given line as this standard form lets us read off $a$ directly, after which the given point of tangency fixes everything else.

    Step-by-Step Solution

    1. The tangent of slope $m$ to $y^2=4ax$ is $y=mx+\frac{a}{m}$; comparing with the given $y=mx+\frac{3}{m}$ gives $\frac{a}{m}=\frac{3}{m}\Rightarrow a=3$.
    2. The point of contact is $\left(\frac{a}{m^2},\frac{2a}{m}\right)=\left(\frac{3}{m^2},\frac{6}{m}\right)$.
    3. This point is $P(3,\beta)$, so $\frac{3}{m^2}=3\Rightarrow m^2=1\Rightarrow m=\pm1$. …

    🔒 Start your 14-day free trial to see the full solution →