JEE Mathematics Ellipse and Hyperbola MCQs Set 05

Answer: (b) 1

Explanation:
1. A chord through \((k,0)\) with end angles \(\alpha,\beta\) satisfies \(\tan\dfrac\alpha2\tan\dfrac\beta2=\dfrac{a-k}{a+k}\).
2. The other point is \((-k,0)\), giving \(\dfrac{a+k}{a-k}\). The product is \(1\).

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Answer: (c) \(\left(\pm2\sqrt3,\pm\dfrac17\right)\)

Explanation:
1. \(P=(4\cos\phi,2\sin\phi)\). The normal \(\dfrac{4x}{\cos\phi}-\dfrac{2y}{\sin\phi}=12\) meets the x-axis at \(Q=(3\cos\phi,0)\). So \(M=\left(\dfrac{7\cos\phi}2,\sin\phi\right)\) and the locus is \(\dfrac{4x^2}{49}+y^2=1\).
2. The latus rectum is \(x=\pm ae=\pm2\sqrt3\). Then \(y^2=1-\dfrac{48}{49}=\dfrac1{49}\), \(y=\pm\dfrac17\).

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Answer: (a) \(\left(\dfrac{2\sqrt{91}}7,\dfrac{3\sqrt{105}}{14}\right)\)

Explanation:
1. The mean of \(4\) and \(3\) is \(\dfrac72\), so \(x^2+y^2=\dfrac{49}4\). With \(9x^2+16y^2=144\): \(9x^2+16\left(\dfrac{49}4-x^2\right)=144\Rightarrow7x^2=52\), \(x^2=\dfrac{52}7\).
2. \(y^2=\dfrac{49}4-\dfrac{52}7=\dfrac{135}{28}\). So \(x=\pm\dfrac{2\sqrt{91}}7\), \(y=\pm\dfrac{3\sqrt{105}}{14}\), and option (a) satisfies both.

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Answer: (b) \(\dfrac{4x^2}{81}+\dfrac{4y^2}{45}=1\)

Explanation:
1. \(\dfrac{2b^2}a=5\) and \(b^2=a^2\left(1-\dfrac49\right)=\dfrac{5a^2}9\). So \(\dfrac{10a}9=5\), \(a=\dfrac92\).
2. \(a^2=\dfrac{81}4\), \(b^2=\dfrac{45}4\): \(\dfrac{4x^2}{81}+\dfrac{4y^2}{45}=1\).

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Answer: (c) \(\sqrt2-1\)

Explanation:
1. \(2ae=\dfrac12\cdot\dfrac{2b^2}a=\dfrac{b^2}a\Rightarrow2a^2e=a^2(1-e^2)\).
2. \(e^2+2e-1=0\Rightarrow e=\sqrt2-1\).

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Answer: (b) an ellipse of eccentricity \(\dfrac{\sqrt3}2\)

Explanation:
1. Let the foot be at \((l\cos\phi,0)\) and the top at \((0,l\sin\phi)\). \(P\) is one third of the length from the wall end: \(P=\left(\dfrac{l\cos\phi}3,\dfrac{2l\sin\phi}3\right)\).
2. Locus: \(\dfrac{x^2}{(l/3)^2}+\dfrac{y^2}{(2l/3)^2}=1\), an ellipse with \(b=\dfrac l3\), \(a=\dfrac{2l}3\): \(e^2=1-\dfrac14=\dfrac34\), \(e=\dfrac{\sqrt3}2\).

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Answer: (d) 3y - 1 = 0

Explanation:
1. Substitute \(x^2=8y\): \(\dfrac{8y}3+y^2=1\Rightarrow3y^2+8y-3=0\Rightarrow y=\dfrac13\) or \(-3\).
2. \(y=-3\) would give \(x^2 \lt 0\), so only \(y=\dfrac13\) gives real points. The line is \(3y-1=0\).

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Answer: (c) 4

Explanation:
1. \(e_1^2=1-\dfrac23=\dfrac13\), so \(e_1=\dfrac1{\sqrt3}\). Then \(e_2=\dfrac{\sqrt3}2\) and \(e_2^2=\dfrac34\).
2. With \(b \lt 4\): \(e_2^2=1-\dfrac{b^2}{16}=\dfrac34\Rightarrow b=2\). The minor axis has length \(2b=4\).

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Answer: (c) 12

Explanation:
1. Continuity from the left: \(a^2=8\). From the right the limit must be finite, so the numerator vanishes at \(x=2\): \(4-b^2=0\Rightarrow b=2\) (\(b>0\)); the limit is \(b(2+b)=8\), consistent.
2. The perpendicular tangents meet on the director circle \(x^2+y^2=a^2+b^2=8+4=12\), so \(r^2=12\).

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Answer: (b) 10

Explanation:
1. \(e=\dfrac35\). \(PS=a-ex=5-\dfrac35x=2\Rightarrow x=5\), so \(P=(5,0)\) is the vertex.
2. The focal chord through the vertex and the focus is the major axis, so \(PQ=2a=10\).

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Answer: (b) 2

Explanation:
1. The focal chord length is \(\dfrac{2ab^2}{a^2\sin^2\theta+b^2\cos^2\theta}\). At \(\theta=\dfrac\pi2\) this gives the latus rectum \(\dfrac{2b^2}a\), confirming \(\lambda=2\).

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Answer: (a) \(\sqrt{\dfrac{17}{20}}\)

Explanation:
1. The tangent condition is \(c^2=a^2m^2-b^2\): \(36=100m^2-49\).
2. \(m^2=\dfrac{85}{100}=\dfrac{17}{20}\).

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Answer: (a) \(\dfrac{x^2}{a^2\sec^2\theta}-\dfrac{y^2}{b^2}=1\)

Explanation:
1. The chord \(PQ\) is \(\dfrac xa\cos\theta-\dfrac yb\sin\phi=\cos\phi\) where \(\phi=\dfrac{\alpha+\beta}2\), i.e. \(\dfrac yb\sin\phi+\cos\phi=\dfrac xa\cos\theta\).
2. As \(\phi\) varies, the envelope is given by \(\left(\dfrac yb\right)^2+1=\dfrac{x^2}{a^2}\cos^2\theta\), i.e. \(\dfrac{x^2}{a^2\sec^2\theta}-\dfrac{y^2}{b^2}=1\).

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Answer: (c) \(\dfrac{(a^2+b^2)^2}{n^2}\)

Explanation:
1. The normal at \((a\sec\phi,b\tan\phi)\) is \(ax\cos\phi+by\cot\phi=a^2+b^2\). Compare with \(lx+my=-n\): \(\dfrac{a\cos\phi}l=\dfrac{b\cot\phi}m=\dfrac{a^2+b^2}{-n}\).
2. Then \(\cos\phi=-\dfrac{l(a^2+b^2)}{an}\) and \(\cot\phi=-\dfrac{m(a^2+b^2)}{bn}\); eliminating \(\phi\) with \(\sec^2\phi-\tan^2\phi=1\) gives \(\dfrac{a^2}{l^2}-\dfrac{b^2}{m^2}=\dfrac{(a^2+b^2)^2}{n^2}\).

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Answer: (c) -1

Explanation:
1. The normal at \(\left(ct,\dfrac ct\right)\) is \(y-\dfrac ct=t^2(x-ct)\). Substituting the point \(\left(ct_1,\dfrac c{t_1}\right)\): \(\dfrac1{t_1}-\dfrac1t=t^2(t_1-t)\).
2. \(\dfrac{t-t_1}{tt_1}=t^2(t_1-t)\Rightarrow-\dfrac1{tt_1}=t^2\Rightarrow t^3t_1=-1\).

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Answer: (c) \(a^2(y^2-x^2)=4x^2y^2\)

Explanation:
1. Let the tangents meet at \((h,k)\); the chord of contact is \(hx-ky=a^2\) with slope \(\dfrac hk\). It is normal at \((x_1,y_1)\), where the normal slope is \(-\dfrac{y_1}{x_1}\), so \(\dfrac hk=-\dfrac{y_1}{x_1}\).
2. With \(hx_1-ky_1=a^2\) this gives \(x_1=\dfrac{a^2}{2h}\), \(y_1=-\dfrac{a^2}{2k}\). Using \(x_1^2-y_1^2=a^2\): \(a^2(k^2-h^2)=4h^2k^2\). The locus is \(a^2(y^2-x^2)=4x^2y^2\).

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Answer: (d) \(\dfrac{-hk}{c^4}\)

Explanation:
1. The normal at \(t\) is \(xt^3-yt-ct^4+c=0\). Through \((h,k)\): \(ct^4-ht^3+kt-c=0\), with roots \(t_1,\dots,t_4\).
2. \(\sum t=\dfrac hc\), \(\prod t=-1\), \(\sum t_it_jt_k=-\dfrac kc\). Then \(\sum x_i=h\), \(\prod x_i=c^4\prod t_i=-c^4\), and \(\sum y_i=c\sum\dfrac1{t_i}=c\cdot\dfrac{-k/c}{-1}=k\). The ratio is \(\dfrac{hk}{-c^4}\).

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Answer: (d) \((5,2\sqrt3)\)

Explanation:
1. The conic is \(\dfrac{x^2}{12}+\dfrac{y^2}{16}=1\): major axis along the y-axis, \(c^2=16-12=4\), foci \((0,\pm2)\).
2. The hyperbola has transverse axis along the y-axis with vertices \((0,\pm2)\): \(a=2\), \(b^2=a^2(e^2-1)=5\): \(\dfrac{y^2}4-\dfrac{x^2}5=1\). Points (a), (b), (c) satisfy it; \((5,2\sqrt3)\) gives \(3-5=-2\ne1\).

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Answer: (d) \(-\dfrac{13}2\)

Explanation:
1. The normal at \(\theta\): \(3x\cos\theta+2y\cot\theta=13\). At \(\phi=\dfrac\pi2-\theta\): \(3x\sin\theta+2y\tan\theta=13\).
2. Multiply the first by \(\sin\theta\) and the second by \(\cos\theta\): \(13\sin\theta-2y\cos\theta=13\cos\theta-2y\sin\theta\). So \(13(\sin\theta-\cos\theta)=2y(\cos\theta-\sin\theta)\), giving \(y=-\dfrac{13}2\) (as \(\theta\ne\phi\)).

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Answer: (a) \(-\dfrac{20}9\)

Explanation:
1. \(a=2\), \(b^2=5\), \(e=\dfrac32\), \(ae=3\). The extremity in the first quadrant is \(\left(3,\dfrac52\right)\).
2. The tangent: \(\dfrac{3x}4-\dfrac{y}2=1\), meeting the axes at \(A=\left(\dfrac43,0\right)\) and \(B=(0,-2)\). \((OA)^2-(OB)^2=\dfrac{16}9-4=-\dfrac{20}9\).

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Answer: (d) \(\dfrac4{x^2}-\dfrac2{y^2}=1\)

Explanation:
1. The tangent at \((2\sec\theta,\sqrt2\tan\theta)\) is \(\dfrac{x\sec\theta}2-\dfrac{y\tan\theta}{\sqrt2}=1\): \(P=(2\cos\theta,0)\), \(Q=(0,-\sqrt2\cot\theta)\). So \(R=(2\cos\theta,-\sqrt2\cot\theta)\).
2. \(\cos\theta=\dfrac x2\), \(\cot^2\theta=\dfrac{y^2}2\). Using \(\cot^2\theta=\dfrac{\cos^2\theta}{1-\cos^2\theta}\): \(\dfrac{y^2}2=\dfrac{x^2}{4-x^2}\), i.e. \(\dfrac4{x^2}-\dfrac2{y^2}=1\).

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Answer: (a) 2

Explanation:
1. From the first, \(t=\dfrac{\sqrt3x-y}{4\sqrt3}\); from the second, \(t(\sqrt3x+y)=4\sqrt3\). Multiply: \((\sqrt3x-y)(\sqrt3x+y)=48\), i.e. \(\dfrac{x^2}{16}-\dfrac{y^2}{48}=1\).
2. \(e^2=1+\dfrac{48}{16}=4\), so \(e=2\).

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Answer: (c) 8

Explanation:
1. The orthocentre of a triangle inscribed in a rectangular hyperbola lies on the hyperbola.
2. \(x\cdot y=4\) with \(x=\dfrac12\) gives \(y=8\).

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Answer: (a) 3

Explanation:
1. The circle \(x^2+y^2=\left(\dfrac a2\right)^2\) is centred at the origin. The point of the hyperbola nearest to the origin is the vertex \((a,0)\), at distance \(a\).
2. So \(f(a,b)=a-\dfrac a2=\dfrac a2\). \(f(4,6)+f(2,3)=2+1=3\).

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