JEE Mathematics Hyperbola MCQs Set 07

Answer: (b) \( x^2-16xy-11y^2-12x+6y+21=0 \)

Explanation:
1. Using \( SP^2=e^2PM^2 \): \( (x-2)^2+(y-1)^2=4\cdot\frac{(x+2y-1)^2}{5} \).
2. Multiplying by 5: \( 5x^2+5y^2-20x-10y+25=4x^2+16y^2+16xy-8x-16y+4 \).
3. Simplifying: \( x^2-16xy-11y^2-12x+6y+21=0 \).

Teacher's Note:
1. Apply the focus-directrix definition with \( e>1 \).

Answer: (b) \( 2a[2-e^2] \)

Explanation:
1. Latus rectum \( =\frac{2b^2}{a}=2a(e^2-1) \).
2. Difference \( =2a-2a(e^2-1)=2a(2-e^2) \).

Teacher's Note:
1. Use \( b^2=a^2(e^2-1) \).

Answer: (d) \( x^2-y^2=0 \)

Explanation:
1. \( xy=1 \) and \( (x-1)(y-3)=3 \) are rectangular hyperbolas with the axes (or shifted axes) as asymptotes, and \( x^2-y^2=5 \) is a rectangular hyperbola.
2. \( x^2-y^2=0 \) factors as \( (x-y)(x+y)=0 \), a pair of straight lines.

Teacher's Note:
1. A zero on the right side of \( x^2-y^2 \) gives lines.

Answer: (a) \( (6,3),(-2,3) \)

Explanation:
1. Completing squares: \( 9(x-2)^2-16(y-3)^2=252+36-144=144 \).
2. So \( \frac{(x-2)^2}{16}-\frac{(y-3)^2}{9}=1 \) with centre \( (2,3) \) and \( a=4 \).
3. Vertices: \( (2\pm4,3)=(6,3) \) and \( (-2,3) \).

Teacher's Note:
1. Shift the centre, then add and subtract a along the transverse axis.

Answer: (c) eccentricity is \( \sqrt{\frac{19}{3}} \)

Explanation:
1. \( 16(x-1)^2-3(y-2)^2=44+16-12=48 \), so \( \frac{(x-1)^2}{3}-\frac{(y-2)^2}{16}=1 \).
2. Here \( a^2=3 \), \( b^2=16 \): transverse axis \( 2\sqrt3 \), conjugate axis 8, centre \( (1,2) \).
3. \( e^2=1+\frac{16}{3}=\frac{19}{3} \), so only (c) is correct.

Teacher's Note:
1. Convert to standard form and check each statement.

Answer: (b) \( \sqrt3 \)

Explanation:
1. The ends of the latus rectum are \( \left(ae,\pm\frac{b^2}{a}\right) \); seen from \( (-ae,0) \) each makes an angle \( 30^\circ \) with the axis.
2. \( \tan30^\circ=\frac{b^2/a}{2ae} \), so \( \frac{b^2}{a}=\frac{2ae}{\sqrt3} \), i.e. \( e^2-1=\frac{2e}{\sqrt3} \).
3. \( \sqrt3e^2-2e-\sqrt3=0 \) gives \( e=\sqrt3 \).

Teacher's Note:
1. Use half of the angle at the other focus.

Answer: (b) \( \frac{y^2}{9}-\frac{x^2}{16}=1 \)

Explanation:
1. The transverse axis is along the y-axis with \( a=3 \) and \( c=5 \).
2. \( b^2=c^2-a^2=16 \), so the equation is \( \frac{y^2}{9}-\frac{x^2}{16}=1 \).

Teacher's Note:
1. For a hyperbola \( c^2=a^2+b^2 \).

Answer: (d) \( x^2+y^2=25 \)

Explanation:
1. The auxiliary circle is drawn on the transverse axis as diameter, with radius a.
2. Here \( a^2=25 \), so the circle is \( x^2+y^2=25 \).

Teacher's Note:
1. Radius of the auxiliary circle is the semi-transverse axis.

Answer: (d) \( 2x+\sqrt3y=25 \)

Explanation:
1. The point lies on the curve: \( 4-3=1 \).
2. The normal at \( (x_1,y_1) \) is \( \frac{a^2x}{x_1}+\frac{b^2y}{y_1}=a^2+b^2 \), i.e. \( \frac{16x}{8}+\frac{9y}{3\sqrt3}=25 \).
3. That is \( 2x+\sqrt3y=25 \).

Teacher's Note:
1. Use the standard normal formula for the hyperbola.

Answer: (c) \( \left(-\frac92,-\frac52\right) \)

Explanation:
1. The hyperbola is \( \frac{x^2}{9}-\frac{y^2}{5}=1 \) and the tangent has \( m=1 \), \( c=2 \) (\( c^2=a^2m^2-b^2=4 \)).
2. The point of contact is \( \left(-\frac{a^2m}{c},-\frac{b^2}{c}\right)=\left(-\frac92,-\frac52\right) \).

Teacher's Note:
1. Use the point of contact formula for \( y=mx+c \).

Answer: (b) \( x^2+y^2=5 \)

Explanation:
1. The hyperbola is \( \frac{x^2}{9}-\frac{y^2}{4}=1 \).
2. The director circle is \( x^2+y^2=a^2-b^2=9-4=5 \).

Teacher's Note:
1. For a hyperbola the director circle is \( x^2+y^2=a^2-b^2 \).

Answer: (c) \( a^2A=b^2B \)

Explanation:
1. The lines are \( y=m_1x \) and \( y=m_2x \) with \( m_1m_2=\frac AB \).
2. Conjugate diameters of the hyperbola satisfy \( m_1m_2=\frac{b^2}{a^2} \).
3. So \( \frac AB=\frac{b^2}{a^2} \), i.e. \( a^2A=b^2B \).

Teacher's Note:
1. Conjugate diameters satisfy \( m_1m_2=b^2/a^2 \).

Answer: (c) \( \lambda<-2 \) or \( \lambda>2 \)

Explanation:
1. Divide both sides by 100: \( \left(x-\frac12\right)^2+\left(y-\frac25\right)^2=\frac{\lambda^2}{100}(3x+4y-1)^2=\frac{\lambda^2}{4}\cdot\left(\frac{3x+4y-1}{5}\right)^2 \).
2. The left side is the squared distance from the point \( \left(\frac12,\frac25\right) \) and the bracket is the distance from the line \( 3x+4y-1=0 \).
3. So the eccentricity is \( e=\frac{|\lambda|}{2} \), and a hyperbola needs \( e>1 \), that is \( |\lambda|>2 \).

Teacher's Note:
1. A conic is a hyperbola when its eccentricity exceeds 1.

Answer: (c) \( \frac2{\sqrt3} \)

Explanation:
1. \( 2b=\frac12(2ae) \) gives \( b=\frac{ae}2 \), so \( b^2=a^2(e^2-1)=\frac{a^2e^2}4 \) and \( e^2=\frac43 \).
2. Hence \( e=\frac{2}{\sqrt3} \) (the latus rectum 8 then gives \( a=12 \)).

Teacher's Note:
1. Use \( b^2=a^2(e^2-1) \) with the given relation between b and ae.

Answer: (a) \( (0,\pm2) \) and 9

Explanation:
1. \( \frac{y^2}{4}-\frac{x^2}{9}=1 \), so the transverse axis is along the y-axis with \( a=2 \), \( b=3 \).
2. Vertices \( (0,\pm2) \) and latus rectum \( \frac{2b^2}{a}=\frac{18}{2}=9 \).

Teacher's Note:
1. Identify which variable has the positive sign.

Answer: (c) \( (x^2-y^2)^2=16xy \)

Explanation:
1. \( x+y=2\tan\theta \) and \( x-y=2\sin\theta \), so \( xy=\tan^2\theta-\sin^2\theta=\tan^2\theta\sin^2\theta \).
2. Hence \( xy=\frac{(x+y)^2}{4}\cdot\frac{(x-y)^2}{4} \), that is \( (x^2-y^2)^2=16xy \).

Teacher's Note:
1. Note that \( \tan^2\theta-\sin^2\theta=\tan^2\theta\sin^2\theta \).

Answer: (d) \( \sqrt{1-\frac1k} \)

Explanation:
1. For \( k<0 \) write the curve as \( \frac{y^2}{k^2}-\frac{x^2}{|k|}=1 \), so \( a^2=k^2 \) and \( b^2=-k \).
2. \( e^2=1+\frac{b^2}{a^2}=1-\frac{k}{k^2}=1-\frac1k \).

Teacher's Note:
1. Rewrite with positive denominators.

Answer: (a) \( \frac{8\sqrt2}{\sqrt3} \)

Explanation:
1. \( \frac{x^2}{32/3}-\frac{y^2}{8}=1 \), so \( a^2=\frac{32}{3} \).
2. \( 2a=2\cdot\frac{4\sqrt2}{\sqrt3}=\frac{8\sqrt2}{\sqrt3} \).

Teacher's Note:
1. The transverse axis is \( 2a \), not \( a^2 \).

Answer: (c) \( \frac1{e^2}+\frac1{e'^2}=1 \)

Explanation:
1. For \( \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \): \( e^2=1+\frac{b^2}{a^2} \), so \( \frac1{e^2}=\frac{a^2}{a^2+b^2} \).
2. For the conjugate: \( e'^2=1+\frac{a^2}{b^2} \), so \( \frac1{e'^2}=\frac{b^2}{a^2+b^2} \).
3. Adding gives 1.

Teacher's Note:
1. Write both eccentricities in terms of a and b.

Answer: (b) \( b^2 \)

Explanation:
1. For the tangent \( y=mx+c \) with \( c^2=a^2m^2-b^2 \), the foci \( (\pm ae,0) \) are at distances \( \frac{|mae+c|}{\sqrt{1+m^2}} \) and \( \frac{|-mae+c|}{\sqrt{1+m^2}} \).
2. The product is \( \frac{|c^2-m^2a^2e^2|}{1+m^2}=\frac{|a^2m^2-b^2-m^2(a^2+b^2)|}{1+m^2}=b^2 \).

Teacher's Note:
1. A standard result, same as for the ellipse.

Answer: (a) \( x^2-y^2=32 \)

Explanation:
1. \( 2ae=16 \) and \( e=\sqrt2 \) give \( a=4\sqrt2 \), so \( a^2=32 \).
2. \( b^2=a^2(e^2-1)=32 \), so the equation is \( x^2-y^2=32 \).

Teacher's Note:
1. Eccentricity \( \sqrt2 \) means a rectangular hyperbola.

Answer: (a) \( \frac{x^2}{4}-\frac{y^2}{5}=\frac49 \)

Explanation:
1. \( ae=2 \) and \( e=\frac32 \) give \( a=\frac43 \), so \( a^2=\frac{16}{9} \).
2. \( b^2=a^2(e^2-1)=\frac{16}{9}\cdot\frac54=\frac{20}{9} \).
3. The equation \( \frac{9x^2}{16}-\frac{9y^2}{20}=1 \) is the same as \( \frac{x^2}{4}-\frac{y^2}{5}=\frac49 \).

Teacher's Note:
1. Multiply the final equation through to compare with the options.

Answer: (a) \( \frac{x^2}{4}-\frac{y^2}{5}=1 \)

Explanation:
1. \( a=2 \) and \( c=3 \), so \( b^2=c^2-a^2=5 \).
2. The equation is \( \frac{x^2}{4}-\frac{y^2}{5}=1 \).

Teacher's Note:
1. Use \( b^2=c^2-a^2 \).

Answer: (c) \( \tan^{-1}\left(\frac{\sqrt3}2\right) \)

Explanation:
1. Substituting \( y=2-\sqrt3x \): \( 4-4\sqrt3x+3x^2-x^2=4 \), so \( x=0 \) or \( x=2\sqrt3 \).
2. The points are \( (0,2) \) and \( (2\sqrt3,-4) \). The first line is the y-axis; the second has slope \( -\frac{2}{\sqrt3} \).
3. The angle with the y-axis has \( \tan=\frac{1}{|m|}=\frac{\sqrt3}{2} \).

Teacher's Note:
1. One of the lines is the y-axis, so use the reciprocal of the other slope.

Answer: (d) \( 2p^2=5 \)

Explanation:
1. The hyperbola is \( \frac{x^2}{9}-\frac{y^2}{4}=1 \) and the line is \( y=-x+\sqrt2p \), so \( m=-1 \) and \( c^2=2p^2 \).
2. The tangency condition \( c^2=a^2m^2-b^2 \) gives \( 2p^2=9-4=5 \).

Teacher's Note:
1. Use \( c^2=a^2m^2-b^2 \).

Answer: (a) \( x^2+y^2=2 \)

Explanation:
1. The director circle is \( x^2+y^2=a^2-b^2=3-1=2 \).

Teacher's Note:
1. Perpendicular tangents meet on the director circle.

Answer: (c) 4

Explanation:
1. The normal in slope form (or parametric form) leads to a quartic equation in m.
2. So at most four normals can be drawn from a point.

Teacher's Note:
1. The equation for the normals through a point is a quartic.

Answer: (b) \( xt^3-yt-at^4+a=0 \)

Explanation:
1. The slope of the tangent at \( \left(at,\frac at\right) \) is \( -\frac{y}{x}=-\frac{1}{t^2} \), so the normal has slope \( t^2 \).
2. Normal: \( y-\frac at=t^2(x-at) \), i.e. \( yt-a=xt^3-at^4 \).
3. Hence \( xt^3-yt-at^4+a=0 \).

Teacher's Note:
1. Normal slope is the negative reciprocal of the tangent slope.

Answer: (a) \( -\frac2{\sqrt3} \)

Explanation:
1. A normal of slope m is \( y=mx\mp\frac{(a^2+b^2)m}{\sqrt{a^2-b^2m^2}} \) with \( a^2+b^2=25 \).
2. So \( \frac{25|m|}{\sqrt{16-9m^2}}=\frac{25}{\sqrt3} \), giving \( 3m^2=16-9m^2 \) and \( m^2=\frac43 \).
3. The intercept is positive, so m must be negative: \( m=-\frac{2}{\sqrt3} \).

Teacher's Note:
1. Check the sign of the intercept against the normal formula.

Answer: (d) \( -\frac{81}{16} \)

Explanation:
1. The hyperbola is \( \frac{x^2}{9}-\frac{y^2}{4}=1 \); the chords of contact are \( \frac{x_ix}{9}-\frac{y_iy}{4}=1 \) with slopes \( \frac{4x_i}{9y_i} \).
2. Perpendicular: \( \frac{16x_1x_2}{81y_1y_2}=-1 \).
3. So \( \frac{x_1x_2}{y_1y_2}=-\frac{81}{16} \).

Teacher's Note:
1. Product of slopes is \( -1 \).

Answer: (b) \( 9x^2-8y^2-18x+9=0 \)

Explanation:
1. The chord of contact from \( (h,k) \) is \( hx-ky=9 \); comparing with \( x=9 \) gives \( (h,k)=(1,0) \).
2. The pair of tangents is \( SS_1=T^2 \): \( (x^2-y^2-9)(1-9)=(x-9)^2 \).
3. \( -8x^2+8y^2+72=x^2-18x+81 \), so \( 9x^2-8y^2-18x+9=0 \).

Teacher's Note:
1. First find the external point from the chord of contact.

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

Explanation:
1. For mid-point \( (h,k) \) of a chord of slope 2: \( T=S_1 \) gives slope \( \frac{3h+2}{2k+3}=2 \) (from the coefficients of the equation).
2. So \( 3h+2=4k+6 \), i.e. \( 3h-4k=4 \).
3. The locus is \( 3x-4y=4 \).

Teacher's Note:
1. The mid-points of parallel chords lie on a diameter.

Answer: (b) \( 75x-16y=418 \)

Explanation:
1. Use \( T=S_1 \): \( 25\cdot6x-16\cdot2y=25\cdot36-16\cdot4 \).
2. \( 150x-32y=836 \), i.e. \( 75x-16y=418 \).

Teacher's Note:
1. Use \( T=S_1 \) for a chord with a given mid-point.

Answer: (a) \( x=k,\ y=h \)

Explanation:
1. \( xy-hx-ky=0 \) gives \( (x-k)(y-h)=hk \).
2. The asymptotes are \( x=k \) and \( y=h \).

Teacher's Note:
1. Factor as \( (x-k)(y-h)=hk \).

Answer: (b) \( \frac23 \)

Explanation:
1. The hyperbola is \( \frac{x^2}{2}-y^2=1 \) with \( a^2=2 \), \( b^2=1 \).
2. The product is \( \frac{a^2b^2}{a^2+b^2}=\frac23 \).

Teacher's Note:
1. The product is \( a^2b^2/(a^2+b^2) \).

Answer: (b) 12

Explanation:
1. For \( xy=c^2 \) the transverse axis is \( 2\sqrt2\,c \), and here \( c^2=18 \), so \( c=3\sqrt2 \).
2. Length \( =2\sqrt2\cdot3\sqrt2=12 \).

Teacher's Note:
1. For \( xy=c^2 \) rotate axes by \( 45^\circ \): \( a=\sqrt2c \).