JEE Mathematics Parabola MCQs Set 11

Answer: (b) parabola

Explanation:
1. The second-degree terms form a perfect square: \(9x^2-24xy+16y^2=(3x-4y)^2\), so \(h^2=ab\). The conic is a parabola (non-degenerate, as the linear terms are not proportional to \(3x-4y\)).

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Answer: (a) \(3x-4y=0\)

Explanation:
1. Put \(T=\dfrac{3x-4y}5\), \(U=\dfrac{4x+3y}5\). Then \(20x+15y=25U\) and the equation becomes \(25T^2=25U+60\), i.e. \(T^2=U+\dfrac{12}5\).
2. The axis is \(T=0\), i.e. \(3x-4y=0\).

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Answer: (c) \(\left(-\dfrac{48}{25},-\dfrac{36}{25}\right)\)

Explanation:
1. From \(T^2=U+\dfrac{12}5\), the vertex has \(T=0\) and \(U=-\dfrac{12}5\), i.e. \(3x-4y=0\) and \(4x+3y=-12\).
2. Solving: \(x=-\dfrac{48}{25}\), \(y=-\dfrac{36}{25}\).

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Answer: (a) \(4x+3y+\dfrac{53}4=0\)

Explanation:
1. \(T^2=\left(U+\dfrac{12}5\right)\) means \(T^2=4\cdot\dfrac14\left(U+\dfrac{12}5\right)\), so the focal length is \(\dfrac14\) and the directrix is \(U+\dfrac{12}5=-\dfrac14\), i.e. \(U=-\dfrac{53}{20}\).
2. \(U=\dfrac{4x+3y}5\), so \(4x+3y=-\dfrac{53}4\), i.e. \(4x+3y+\dfrac{53}4=0\).

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Answer: (a) \(x^2+y^2-4ax=0\)

Explanation:
1. Chords subtending a right angle at the vertex all pass through the fixed point \((4a,0)\).
2. The foot of the perpendicular from \(O\) to a chord through \((4a,0)\) sees the segment from \(O\) to \((4a,0)\) at a right angle, so it lies on the circle with that segment as diameter: \(x^2+y^2-4ax=0\).

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Answer: (c) \(9x^2=4by\)

Explanation:
1. A double ordinate has ends \((\pm2bt,bt^2)\). A trisection point is \((h,k)=\left(\dfrac{2bt}3,\,bt^2\right)\).
2. \(t=\dfrac{3h}{2b}\) gives \(k=\dfrac{9h^2}{4b}\), i.e. \(9x^2=4by\).

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Answer: (d) \(xy=\dfrac{105}{64}\)

Explanation:
1. The vertex of \(y=px^2+qx+r\) is \(\left(-\dfrac q{2p},\,r-\dfrac{q^2}{4p}\right)\). Here \(x=-\dfrac3{4a}\) and \(y=-2a-\dfrac{3a}{16}=-\dfrac{35a}{16}\).
2. \(xy=\dfrac{3}{4a}\cdot\dfrac{35a}{16}=\dfrac{105}{64}\).

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Answer: (c) \(\left(\dfrac52,\dfrac72\right)\)

Explanation:
1. The given line is the chord of contact (polar) of the required point \((h,k)\): \(ky=2(x+h)\), i.e. \(4x-2ky+4h=0\). Compare with \(4x-7y+10=0\): \(k=\dfrac72\), \(h=\dfrac52\).
2. Check: \(A=(1,2)\), \(B=\left(\dfrac{25}4,5\right)\); the tangents \(y=x+1\) and \(5y=2x+\dfrac{25}2\) meet at \(\left(\dfrac52,\dfrac72\right)\).

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Answer: (c) \(y=2(x+1)\)

Explanation:
1. \(AB\) is the chord of contact of \(P(h,k)\): \(ky=2(x+h)\). It passes through \((1,1)\): \(k=2(1+h)\).
2. The locus is \(y=2(x+1)\).

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Answer: (b) G.P.

Explanation:
1. Let \(B=(at_1^2,2at_1)\), \(C=(at_2^2,2at_2)\), so \(A=(at_1t_2,a(t_1+t_2))\). For the tangent \(ty=x+at^2\), the perpendicular distances are proportional to \(a(t-t_1)(t-t_2)\), \(a(t-t_1)^2\), \(a(t-t_2)^2\) (all divided by \(\sqrt{1+t^2}\)).
2. So \(p_1=k(t-t_1)(t-t_2)\), \(p_2=k(t-t_1)^2\), \(p_3=k(t-t_2)^2\) and \(p_1^2=p_2p_3\). Hence \(p_2,p_1,p_3\) are in G.P.

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Answer: (b) \((x+1)(y-1)^2+4=0\)

Explanation:
1. The parabola is \((y-1)^2=4(x-1)\): vertex \((1,1)\), directrix \(x=0\). Take \(P=(t^2+1,2t+1)\); the tangent is \(t(y-1)=(x-1)+t^2\), meeting \(x=0\) at \(Q=\left(0,1+\dfrac{t^2-1}t\right)\).
2. \(R\) divides \(QP\) externally in the ratio \(1:2\): \(R=2Q-P=\left(-(t^2+1),\ \dfrac{t-2}t\right)\).
3. Then \(x+1=-t^2\) and \(y-1=-\dfrac2t\), so \((x+1)(y-1)^2=-4\), i.e. \((x+1)(y-1)^2+4=0\).

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Answer: (a) \(x=3\left[7\left(\dfrac y{18}\right)^{2/3}+2\right]\)

Explanation:
1. \(a=3\); ordinates \(6t_1,6t_2\) in ratio \(1:2\) give \(t_2=2t_1\). The normals at \(t_1,t_2\) meet at \(x=a(t_1^2+t_1t_2+t_2^2+2)\), \(y=-at_1t_2(t_1+t_2)\).
2. \(x=3(7t_1^2+2)\) and \(y=-18t_1^3\), so \(t_1^2=\left(\dfrac y{18}\right)^{2/3}\) and \(x=3\left[7\left(\dfrac y{18}\right)^{2/3}+2\right]\).

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Answer: (b) \(4\sqrt6\,a\)

Explanation:
1. The normal at \(t\) meets the curve again at \(t_1=-t-\dfrac2t\), so \(t_1^2=\left(t+\dfrac2t\right)^2\ge8\).
2. Distance from the vertex: \(a|t_1|\sqrt{t_1^2+4}\), increasing in \(t_1^2\); its minimum is at \(t_1^2=8\): \(a\sqrt8\sqrt{12}=4\sqrt6\,a\).

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Answer: (a) \(2xm^2-ym^3=4a(2+m^2)\)

Explanation:
1. A chord of slope \(m\) joins \(t_1,t_2\) with \(t_1+t_2=\dfrac2m\). The normals at \(t_1,t_2\) meet at \(h=a(s^2-p+2)\), \(k=-aps\) where \(s=t_1+t_2,\ p=t_1t_2\).
2. Eliminating \(p\): \(h=as^2+\dfrac ks+2a\). With \(s=\dfrac2m\): \(2m^2h-km^3=4a(2+m^2)\).

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Answer: (c) \(a\sqrt{1+t^2}\)

Explanation:
1. The circle on \(SP\) as diameter meets the normal at \(P\) and at the foot \(F\) of the perpendicular from the focus \(S\) to the normal. The intercept is \(PF\).
2. The normal has direction \((1,-t)\); \(PF\) is the projection of \(\overrightarrow{PS}=(a-at^2,-2at)\) on it: \(\dfrac{a(1-t^2)+2at^2}{\sqrt{1+t^2}}=a\sqrt{1+t^2}\).

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Answer: (a) \(\left(\dfrac nl,-\dfrac{2am}l\right)\)

Explanation:
1. The given line is the chord of contact of the required point \((h,k)\): \(ky=2a(x+h)\), i.e. \(2ax-ky+2ah=0\).
2. Comparing with \(lx+my+n=0\): \(\dfrac{2a}l=\dfrac{-k}m=\dfrac{2ah}n\), so \(h=\dfrac nl\) and \(k=-\dfrac{2am}l\).

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Answer: (c) the directrix

Explanation:
1. The polar of \((h,k)\) is \(ky=2a(x+h)\). It passes through the focus \((a,0)\) iff \(a+h=0\).
2. So the locus is \(x=-a\), the directrix.

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Answer: (d) \(\sqrt2\)

Explanation:
1. Interchanging \(x\) and \(y\) maps one equation to the other, so the parabolas are mirror images in \(y=x\). They do not meet (the line \(y=x\) meets neither: \(x^2-12x+40=0\) has no real roots).
2. The closest points have parallel tangents with slope \(1\). On \(x^2-8x-4y+40=0\), slope \(\dfrac{x-4}2=1\) gives \(P=(6,7)\); its mirror is \(Q=(7,6)\). \(PQ=\sqrt2\).

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Answer: (a) \(\left(-\infty,-\sqrt{\dfrac{\sqrt2-1}2}\right)\cup\left(\sqrt{\dfrac{\sqrt2-1}2},\infty\right)\)

Explanation:
1. A tangent of slope \(m\) to the parabola is \(y=mx+\dfrac1m\). It is a chord of the circle if its distance from the origin is less than \(2\): \(\dfrac1{|m|\sqrt{1+m^2}} \lt 2\).
2. \(4m^2(1+m^2)>1\Rightarrow4m^4+4m^2-1>0\Rightarrow m^2>\dfrac{\sqrt2-1}2\).

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Answer: (b) \(\left(10+3\sqrt[3]2,\infty\right)\)

Explanation:
1. Every normal of the circle passes through its centre \(\left(1,\dfrac h2\right)\), so the common normals are the normals to the parabola through that point.
2. With \(X=x-2\), \(Y=y-3\) the parabola is \(X^2=4Y\) and a normal has the form \(X=m(Y)-2m-m^3\) (slope parameter \(m\)). Passing through \(\left(-1,\dfrac h2-3\right)\) gives \(m^3-\left(\dfrac h2-5\right)m-1=0\).
3. Three distinct real roots need \(4\left(5-\dfrac h2\right)^3+27 \lt 0\), i.e. \(\dfrac h2-5>\dfrac{3}{\sqrt[3]4}\), so \(h>10+3\sqrt[3]2\approx13.78\).

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Answer: (b) \(9y^2=-4x\)

Explanation:
1. \(P=(-t^2,2t)\) with \(t>0\) and \(Q=(-t^2,-2t)\). \(R\) divides \(PQ\) in the ratio \(2:1\): \(R=\left(-t^2,\dfrac{2(-2t)+1\cdot2t}3\right)=\left(-t^2,-\dfrac{2t}3\right)\).
2. \(y^2=\dfrac{4t^2}9=-\dfrac{4x}9\), so \(9y^2=-4x\).

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Answer: (c) \(L_1 \lt L_2\)

Explanation:
1. Substitute \(y^2=8x\): \(x^2+8x-9=0\Rightarrow x=1\) (\(x=-9\) rejected), \(y=\pm2\sqrt2\). So \(L_1=4\sqrt2\approx5.66\).
2. \(L_2=4a=8\). Hence \(L_1 \lt L_2\).

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Answer: (a) \(\dfrac{\sqrt5}2\)

Explanation:
1. The focus is \(\left(\dfrac32,0\right)\). The chord is \(y=m\left(x-\dfrac32\right)\), at distance \(\dfrac{3|m|/2}{\sqrt{1+m^2}}\) from the origin.
2. \(\dfrac{9m^2}{4(1+m^2)}=\dfrac54\Rightarrow9m^2=5+5m^2\Rightarrow m^2=\dfrac54\), so \(m=\pm\dfrac{\sqrt5}2\).

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Answer: (b) Statement 1 is true; Statement 2 is true; Statement 2 is a correct explanation for Statement 1

Explanation:
1. Statement 1: \(x=2+2y\) in \(y^2+2x=0\) gives \((y+2)^2=0\), a single point \((-2,-2)\): true.
2. Statement 2: for \(y^2=-4a'x\) with \(a'=\tfrac12\), the tangent \(y=mx-\dfrac{a'}m\) touches at \(\left(-\dfrac{a'}{m^2},-\dfrac{2a'}m\right)\): true.
3. The line in Statement 1 is \(y=\tfrac12x-1\), the case \(m=\tfrac12\) of Statement 2, so Statement 2 explains Statement 1.

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Answer: (b) (3, 0)

Explanation:
1. The ends of the latus rectum have parameters \(t=\pm1\). The normals meet at \(x=a(t_1^2+t_1t_2+t_2^2+2)=1-1+1+2=3\) and \(y=-at_1t_2(t_1+t_2)=0\).
2. The point is \((3,0)\).

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Answer: (c) 16

Explanation:
1. Let \(P=(t^2,2t)\) with \(1\le t\le2\). The tangent meets the x-axis at \((-t^2,0)\); the ordinate's foot is \((t^2,0)\). Base \(=2t^2\), height \(=2t\).
2. Area \(=\dfrac12\cdot2t^2\cdot2t=2t^3\), maximum at \(t=2\): \(\lambda=16\).

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Answer: (a) 12

Explanation:
1. The normal at \(t=1\) meets the curve again at \(t'=-t-\dfrac2t=-3\): \((9,-6)\). The normal at \(t=-1\) meets it again at \(t'=3\): \((9,6)\).
2. \(AA'=12\).

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Answer: (d) 2

Explanation:
1. The orthocentre of the triangle is \(\left(-a(1+t_1t_2+t_2t_3+t_3t_1),\ a(t_1+t_2+t_3+t_1t_2t_3)\right)\).
2. Setting this equal to \((a,0)\): \(1+\sum t_it_j=-1\Rightarrow\sum t_it_j=-2\), so the absolute value is \(2\).

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Answer: (a) 6

Explanation:
1. The focal distance of \((t^2,2t)\) is \(1+t^2\): \(t_1^2=3\), \(t_2^2=8\). The tangents meet at \((t_1t_2,\,t_1+t_2)\).
2. Distance squared from \((1,0)\): \((t_1t_2-1)^2+(t_1+t_2)^2=t_1^2t_2^2+t_1^2+t_2^2+1=24+3+8+1=36\). Distance \(=6\).

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