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Parabola question

2014 · Shift 2 · Q32
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  5. /2014 · Shift 2 · Q32

Parabola question

2014 · Shift 2 · Q32

JEE AdvancedMathematicsParabolaMCQ+3 / −1
Let a,r,s,ta, r, s, ta,r,s,t be nonzero real numbers. Let P  (at2,2at),  Q,   R  (ar2,2ar)P\,\,\left( {a{t^2},2at} \right),\,\,Q,\,\,\,R\,\,\left( {a{r^2},2ar} \right)P(at2,2at),Q,R(ar2,2ar) and S  (as2,2as)S\,\,\left( {a{s^2},2as} \right)S(as2,2as) be distinct points on the parabola y2=4ax{y^2} = 4axy2=4ax. Suppose that PQPQPQ is the focal chord and lines QRQRQR and PKPKPK are parallel, where KKK is the point (2a,0)(2a,0)(2a,0) The value of rrr is
  1. A
    −1t- {1 \over t}−t1​
  2. B
    t2+1t{{{t^2} + 1} \over t}tt2+1​
  3. C
    1t{1 \over t}t1​
  4. D
    t2−1t{{{t^2} - 1} \over t}tt2−1​
View written solutionFree

Correct answer: D

  1. Parametric form of the parabola

For the parabola y2=4axy^2=4axy2=4ax, a general point is

(at2,2at).(at^2,2at).(at2,2at).

So the given points are:

  • P(at2,2at)P(at^2,2at)P(at2,2at)
  • R(ar2,2ar)R(ar^2,2ar)R(ar2,2ar)
  • QQQ is also on the parabola, say Q(aq2,2aq)Q(aq^2,2aq)Q(aq2,2aq) for some parameter qqq.
  1. Use the fact that PQPQPQ is a focal chord

For the parabola y2=4axy^2=4axy2=4ax, if two points with parameters t1,t2t_1,t_2t1​,t2​ form a focal chord, then

t1t2=−1.t_1t_2=-1.t1​t2​=−1.

Here PPP has parameter ttt, and QQQ has parameter qqq. Therefore,

tq=−1⇒q=−1t.tq=-1 \quad \Rightarrow \quad q=-\frac1t.tq=−1⇒q=−t1​.

Hence

Q=(a1t2,−2at).Q=\left(a\frac{1}{t^2},-\frac{2a}{t}\right).Q=(at21​,−t2a​).
  1. Slope of PKPKPK

Given K=(2a,0)K=(2a,0)K=(2a,0) and P=(at2,2at)P=(at^2,2at)P=(at2,2at).

Slope of PKPKPK is

mPK=0−2at2a−at2=−2ata(2−t2)=−2t2−t2=2tt2−2.m_{PK}=\frac{0-2at}{2a-at^2} =\frac{-2at}{a(2-t^2)} =\frac{-2t}{2-t^2} =\frac{2t}{t^2-2}.mPK​=2a−at20−2at​=a(2−t2)−2at​=2−t2−2t​=t2−22t​.
  1. Slope of QRQRQR

Now

Q=(at2,−2at),R=(ar2,2ar).Q=\left(\frac{a}{t^2},-\frac{2a}{t}\right), \qquad R=(ar^2,2ar).Q=(t2a​,−t2a​),R=(ar2,2ar).

So

mQR=2ar−(−2at)ar2−at2=2a(r+1t)a(r2−1t2).m_{QR}=\frac{2ar-\left(-\frac{2a}{t}\right)}{ar^2-\frac{a}{t^2}} =\frac{2a\left(r+\frac1t\right)}{a\left(r^2-\frac1{t^2}\right)}.mQR​=ar2−t2a​2ar−(−t2a​)​=a(r2−t21​)2a(r+t1​)​.

Factor the denominator:

r2−1t2=(r−1t)(r+1t).r^2-\frac1{t^2}=\left(r-\frac1t\right)\left(r+\frac1t\right).r2−t21​=(r−t1​)(r+t1​).

Thus,

mQR=2(r+1t)(r−1t)(r+1t)=2r−1t.m_{QR}=\frac{2\left(r+\frac1t\right)}{\left(r-\frac1t\right)\left(r+\frac1t\right)} =\frac{2}{r-\frac1t}.mQR​=(r−t1​)(r+t1​)2(r+t1​)​=r−t1​2​.
  1. Use the condition QR∥PKQR \parallel PKQR∥PK

Since QRQRQR and PKPKPK are parallel,

mQR=mPK.m_{QR}=m_{PK}.mQR​=mPK​.

So

2r−1t=2tt2−2.\frac{2}{r-\frac1t}=\frac{2t}{t^2-2}.r−t1​2​=t2−22t​.

Cancel 222:

1r−1t=tt2−2.\frac{1}{r-\frac1t}=\frac{t}{t^2-2}.r−t1​1​=t2−2t​.

Invert:

r−1t=t2−2t.r-\frac1t=\frac{t^2-2}{t}.r−t1​=tt2−2​.

Hence

r=1t+t2−2t=t2−1t.r=\frac1t+\frac{t^2-2}{t}=\frac{t^2-1}{t}.r=t1​+tt2−2​=tt2−1​.
  1. Match with the options
r=t2−1tr=\frac{t^2-1}{t}r=tt2−1​

which is Option D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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