Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Parabola question

2009 · Shift 2 · Q35
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Mathematics
  4. /Parabola
  5. /2009 · Shift 2 · Q35

Parabola question

2009 · Shift 2 · Q35

JEE AdvancedMathematicsParabolaMultiple correct+4 / −2
The tangent PTPTPT and the normal PNPNPN to the parabola y2=4ax{y^2} = 4axy2=4ax at a point PPP on it meet its axis at points TTT and NNN, respectively. The locus of the centroid of the triangle PTNPTNPTN is a parabola whose
  1. A
    vertex is (2a3,0)\left( {{{2a} \over 3},0} \right)(32a​,0)
  2. B
    directrix is x=0x=0x=0
  3. C
    latus rectum is 2a3{{{2a} \over 3}}32a​
  4. D
    focus is (a,0)(a, 0)(a,0)
View written solutionFree

Correct answer: A, D

  1. Parametrize the point on the parabola

For the parabola y2=4ax,y^2=4ax,y2=4ax, a general point on it is P(at2,2at).P(at^2,2at).P(at2,2at).

Its axis is the xxx-axis.

  1. Equation of tangent at P(at2,2at)P(at^2,2at)P(at2,2at)

For y2=4axy^2=4axy2=4ax, the tangent at parameter ttt is ty=x+at2.ty=x+at^2.ty=x+at2.

To find where it meets the axis, put y=0y=0y=0: 0=x+at2  ⟹  x=−at2.0=x+at^2 \implies x=-at^2.0=x+at2⟹x=−at2.

So, T(−at2,0).T(-at^2,0).T(−at2,0).

  1. Equation of normal at P(at2,2at)P(at^2,2at)P(at2,2at)

For y2=4axy^2=4axy2=4ax, differentiating: 2ydydx=4a  ⟹  dydx=2ay.2y\frac{dy}{dx}=4a \implies \frac{dy}{dx}=\frac{2a}{y}.2ydxdy​=4a⟹dxdy​=y2a​.

At P(at2,2at)P(at^2,2at)P(at2,2at), mtangent=2a2at=1t.m_{\text{tangent}}=\frac{2a}{2at}=\frac{1}{t}.mtangent​=2at2a​=t1​.

Hence slope of normal is mnormal=−t.m_{\text{normal}}=-t.mnormal​=−t.

Equation of normal through P(at2,2at)P(at^2,2at)P(at2,2at): y−2at=−t(x−at2).y-2at=-t(x-at^2).y−2at=−t(x−at2).

To find its intersection with the axis, put y=0y=0y=0: −2at=−t(x−at2).-2at=-t(x-at^2).−2at=−t(x−at2).

If t≠0t\neq 0t=0, divide by −t-t−t: 2a=x−at2  ⟹  x=2a+at2.2a=x-at^2 \implies x=2a+at^2.2a=x−at2⟹x=2a+at2.

So, N(2a+at2,0).N(2a+at^2,0).N(2a+at2,0).

(For t=0t=0t=0, the same point is obtained in limiting form, so the locus remains valid.)

  1. Centroid of triangle PTNPTNPTN

Let centroid be G(h,k)G(h,k)G(h,k).

Using coordinates: P(at2,2at),T(−at2,0),N(2a+at2,0).P(at^2,2at),\quad T(-at^2,0),\quad N(2a+at^2,0).P(at2,2at),T(−at2,0),N(2a+at2,0).

Then h=at2+(−at2)+(2a+at2)3=a(t2+2)3,h=\frac{at^2+(-at^2)+(2a+at^2)}{3}=\frac{a(t^2+2)}{3},h=3at2+(−at2)+(2a+at2)​=3a(t2+2)​, k=2at+0+03=2at3.k=\frac{2at+0+0}{3}=\frac{2at}{3}.k=32at+0+0​=32at​.

So, x=a(t2+2)3,y=2at3.x=\frac{a(t^2+2)}{3}, \qquad y=\frac{2at}{3}.x=3a(t2+2)​,y=32at​.

  1. Eliminate the parameter

From y=2at3,y=\frac{2at}{3},y=32at​, we get t=3y2a.t=\frac{3y}{2a}.t=2a3y​.

Substitute into xxx:

=\frac{3y^2}{4a}+\frac{2a}{3}.$$ Hence $$x-\frac{2a}{3}=\frac{3y^2}{4a}.$$ Rearranging, $$y^2=\frac{4a}{3}\left(x-\frac{2a}{3}\right).$$ This is a parabola of the form $$y^2=4A(x-h),$$ with $$h=\frac{2a}{3}, \qquad A=\frac{a}{3}.$$ --- 6. **Read geometric elements of the locus** For $$y^2=4A(x-h),$$ - vertex is $(h,0)$, - focus is $(h+A,0)$, - directrix is $x=h-A$, - length of latus rectum is $4A$. Here $A=\frac{a}{3}$ and $h=\frac{2a}{3}$. Thus: - **Vertex** $=\left(\frac{2a}{3},0\right)$ - **Focus** $=\left(\frac{2a}{3}+\frac{a}{3},0\right)=(a,0)$ - **Directrix** $=x=\frac{2a}{3}-\frac{a}{3}=\frac{a}{3}$ - **Latus rectum** $=4\cdot \frac{a}{3}=\frac{4a}{3}$ --- 7. **Check options** - **A:** vertex is $\left(\frac{2a}{3},0\right)$ ✔️ - **B:** directrix is $x=0$ ✖️ (actual directrix is $x=\frac{a}{3}$) - **C:** latus rectum is $\frac{2a}{3}$ ✖️ (actual latus rectum is $\frac{4a}{3}$) - **D:** focus is $(a,0)$ ✔️ --- 8. **Final answer** The correct options are $$\boxed{A, D}.$$
PreviousNext

More from Parabola

  • The locus of the orthocentre of the triangle formed by the lines (1+p)x−py+p(1+p)=0,(1+q)x−qy+q(1+q)=0 and y=0, where peq, is :2009 · MCQ
  • The tangent to the curve y=ex drawn at the point (c,ec) intersects the line joining the points (c−1,ec−1) and (c+1,ec+1)2007 · MCQ
  • Consider the circle x2+y2=9 and the parabola y2=8x. They intersect at P and Q in the first and the fourth quadrants, respectively. Tangents to the circle at P and Q intersect the x-axis at R and tangents to the parabola…2007 · MCQ
  • Consider the circle x2+y2=9 and the parabola y2=8x. They intersect at P and Q in the first and the fourth quadrants, respectively. Tangents to the circle at P and Q intersect the x-axis at R and tangents to the parabola…2007 · MCQ
  • Consider the circle x2+y2=9 and the parabola y2=8x. They intersect at P and Q in the first and the fourth quadrants, respectively. Tangents to the circle at P and Q intersect the x-axis at R and tangents to the parabola…2007 · MCQ
  • Let S denote the locus of the mid-points of those chords of the parabola y2=x, such that the area of the region enclosed between the parabola and the chord is 34​. Let R denote the region lying in the first…2025 · Multiple correct
  • Let A1​,B1​,C1​ be three points in the xy-plane. Suppose that the lines A1​C1​ and B1​C1​ are tangents to the curve y2=8x at A1​ and B1​, respectively. If O=(0,0) and C1​=(−4,0), then which of the following…2024 · Multiple correct
  • A normal with slope 6​1​ is drawn from the point (0,−α) to the parabola x2=−4ay, where a>0. Let L be the line passing through (0,−α) and parallel to the directrix of the parabola. Suppose that L…2024 · Numerical