JEE Mathematics Circle and Conic Section MCQs Set 27

Answer: (b) 75

Explanation:
1. Centre \( A(1,2) \), radius \( 5 \). The tangent at \( B \) is \( y=7 \); the tangent at \( D \) is \( 3(x-1)-4(y-2)=25 \).
2. Solving, \( C=(16,7) \), so \( BC=15 \).
3. \( ABCD \) is made of two congruent right triangles \( ABC \) and \( ADC \): area \( =2\cdot\frac12\cdot5\cdot15=75 \).

Teacher's Note:
1. The radius is perpendicular to the tangent at the point of contact, so \( ABCD \) splits into two right triangles.
2. Find the tangents' meeting point first, then add the two areas.

Answer: (d) \( 3a^2+2ab+3b^2=0 \)

Explanation:
1. If the angle between the lines is \( \theta \), the sectors have angles \( \theta,\pi-\theta,\theta,\pi-\theta \). One is thrice another, so \( \pi-\theta=3\theta \) and \( \theta=\frac\pi4 \).
2. For \( ax^2+2hxy+by^2=0 \): \( \tan\theta=\frac{2\sqrt{h^2-ab}}{|a+b|} \) with \( h=a+b \). Setting \( \tan\theta=1 \): \( 4\{(a+b)^2-ab\}=(a+b)^2 \).
3. This gives \( 3(a+b)^2=4ab \), i.e. \( 3a^2+2ab+3b^2=0 \).

Teacher's Note:
1. Sector areas are proportional to their central angles.
2. Use the angle formula for a homogeneous pair of lines.

Answer: (c) \( \sin^{-1}\left(\frac{b}{am}\right) \)

Explanation:
1. Tangent at \( \varphi \): \( \frac{x\sec\varphi}{a}-\frac{y\tan\varphi}{b}=1 \), so its slope is \( m=\frac{b\sec\varphi}{a\tan\varphi}=\frac{b}{a\sin\varphi} \).
2. Hence \( \sin\varphi=\frac{b}{am} \) and \( \varphi=\sin^{-1}\left(\frac{b}{am}\right) \).

Teacher's Note:
1. Write the tangent in parametric form and compare slopes.
2. \( \sec\varphi/\tan\varphi=1/\sin\varphi \) simplifies the slope.

Answer: (d) \( x^2+y^2+32x+4y+235=0 \)

Explanation:
1. Centre \( (-8,12) \), \( r^2=64+144-183=25 \). The image has the same radius.
2. \( \frac{4(-8)+7(12)+13}{65}=1 \), so the image of the centre is \( (-8,12)-2(1)(4,7)=(-16,-2) \).
3. Image circle: \( x^2+y^2+32x+4y+256+4-25=0 \), i.e. \( x^2+y^2+32x+4y+235=0 \).

Teacher's Note:
1. Reflect only the centre; the radius does not change.
2. Use the reflection formula \( (x',y')=(x_1,y_1)-2\frac{ax_1+by_1+c}{a^2+b^2}(a,b) \).

Answer: (c) 8

Explanation:
1. \( a=4,\ b=2\sqrt2,\ c=\sqrt{16-8}=2\sqrt2 \). Base \( S_1S_2=2c=4\sqrt2 \).
2. Height is maximum when \( A \) is an end of the minor axis: \( b=2\sqrt2 \).
3. Maximum area \( =\frac12\cdot4\sqrt2\cdot2\sqrt2=8 \).

Teacher's Note:
1. The base is fixed, so the area is largest when the height is largest.
2. The greatest distance of a point of the ellipse from the major axis is \( b \).

Answer: (b) G.P.

Explanation:
1. Distances from the origin to the centres are \( |\lambda_i| \), which are in G.P.
2. For a point \( (h,k) \) on \( x^2+y^2=c^2 \), the tangent length squared is \( h^2+k^2-2\lambda_ih-c^2=-2\lambda_ih \).
3. So the lengths are proportional to \( \sqrt{\lambda_i} \), and square roots of terms of a G.P. are also in G.P.

Teacher's Note:
1. Use \( t^2=S_1 \) for the tangent length.
2. Square roots of a G.P. form a G.P. with the square-rooted ratio.

Answer: (c) \( a^2-b^2=2c^2 \)

Explanation:
1. Orthogonal intersection of an ellipse and a hyperbola requires that they are confocal.
2. Foci of the ellipse: \( a^2-b^2 \); foci of \( \frac{x^2}{c^2}-\frac{y^2}{c^2}=1 \): \( c^2+c^2=2c^2 \). Hence \( a^2-b^2=2c^2 \).

Teacher's Note:
1. Confocal conics intersect orthogonally.
2. Compare the squares of the focal distances \( (ae)^2 \).

Answer: (d) Both (b) and (c)

Explanation:
1. Let the tangent slope be \( m \): \( \left|\frac{m-3}{1+3m}\right|=1 \Rightarrow m=-2 \) or \( m=\frac12 \).
2. Tangent \( y=mx+\frac2m \): for \( m=-2 \): \( 2x+y+1=0 \); for \( m=\frac12 \): \( x-2y+8=0 \). Both (b) and (c) hold.

Teacher's Note:
1. Use the angle formula between two lines to find the possible slopes.
2. Tangent to \( y^2=4ax \) with slope \( m \) is \( y=mx+\frac am \).

Answer: (c) \( \frac\pi2 \)

Explanation:
1. For the point \( (-4a,k) \), the chord of contact is \( ky=2a(x-4a) \), which always passes through \( (4a,0) \).
2. A chord through \( (4a,0) \) meets the parabola at \( t_1,t_2 \) with \( t_1t_2=-4 \), the condition for the chord to subtend a right angle at the vertex. The angle is \( \frac\pi2 \).

Teacher's Note:
1. The chord of contact passes through a fixed point on the axis.
2. Chords through \( (4a,0) \) subtend \( 90^\circ \) at the vertex.

Answer: (a) \( 4a^2\left(1-\frac{b^2}{d^2}\right) \)

Explanation:
1. \( (PF_1-PF_2)^2=(PF_1+PF_2)^2-4PF_1\cdot PF_2=4a^2-4PF_1\cdot PF_2 \).
2. By Apollonius, \( PF_1\cdot PF_2=CD^2 \), where \( CD \) is the conjugate semi-diameter, and \( d\cdot CD=ab \). So \( PF_1\cdot PF_2=\frac{a^2b^2}{d^2} \).
3. Hence \( (PF_1-PF_2)^2=4a^2\left(1-\frac{b^2}{d^2}\right) \). A numerical check with \( a=5,b=3 \) confirms it.

Teacher's Note:
1. Use \( (u-v)^2=(u+v)^2-4uv \) with \( u+v=2a \).
2. The product of the focal distances equals \( \frac{a^2b^2}{d^2} \).

Answer: (b) \( y^2=\left(m^{1/2}+m^{-1/2}\right)^2ax \)

Explanation:
1. Points \( t_1,t_2 \) of the same sign with \( \frac{t_1^2}{t_2^2}=m^2 \) give \( t_1=mt_2 \). Tangents meet at \( (at_1t_2,a(t_1+t_2)) \).
2. \( x=amt_2^2 \), \( y=a(m+1)t_2 \), so \( y^2=\frac{(m+1)^2}{m}ax=\left(m^{1/2}+m^{-1/2}\right)^2ax \).

Teacher's Note:
1. Use the parametric form \( (at^2,2at) \) and the intersection point of tangents.
2. Same-side points have \( t_1,t_2 \) of the same sign.

Answer: (b) \( (-1,2),\left(\frac35,\frac{14}5\right) \)

Explanation:
1. Coaxial family \( S_1+\lambda S_2=0 \) has centre \( \left(\frac{1-3\lambda}{1+\lambda},\frac{3+\lambda}{1+\lambda}\right) \) and constant \( \frac{9+\lambda}{1+\lambda} \).
2. Radius zero: \( g^2+f^2-c=0 \) gives \( \lambda=\frac19 \) or \( 1 \). The centres are \( \left(\frac35,\frac{14}5\right) \) and \( (-1,2) \).

Teacher's Note:
1. Limiting points are circles of zero radius in the coaxial family.
2. Solve \( g^2+f^2-c=0 \) for \( \lambda \) and find the corresponding centres.

Answer: (a) \( (y+y_1)^2=8a(x+x_1) \)

Explanation:
1. If \( Q=(x,y) \), then \( R=\left(\frac{x+x_1}2,\frac{y+y_1}2\right) \) lies on the parabola.
2. \( \left(\frac{y+y_1}2\right)^2=4a\cdot\frac{x+x_1}2\Rightarrow(y+y_1)^2=8a(x+x_1) \).

Teacher's Note:
1. Use the midpoint formula and substitute into the parabola.
2. The locus is a parabola of double the latus rectum, shifted.

Answer: (b) G.P.

Explanation:
1. With \( P(at_1^2,2at_1) \), \( Q(at_2^2,2at_2) \): \( SP=a(1+t_1^2) \), \( SQ=a(1+t_2^2) \), \( T=(at_1t_2,a(t_1+t_2)) \).
2. \( ST^2=a^2(t_1t_2-1)^2+a^2(t_1+t_2)^2=a^2(1+t_1^2)(1+t_2^2)=SP\cdot SQ \). So they are in G.P.

Teacher's Note:
1. Use focal distance \( a(1+t^2) \).
2. \( (t_1t_2-1)^2+(t_1+t_2)^2=(1+t_1^2)(1+t_2^2) \).

Answer: (c) 10

Explanation:
1. \( \frac{x^2}{25}+\frac{y^2}{16}=1 \): \( a=5,b=4,c=3 \), so \( F_1,F_2 \) are the foci.
2. \( PF_1+PF_2=2a=10 \).

Teacher's Note:
1. Verify the given points are the foci using \( c^2=a^2-b^2 \).
2. The sum of focal distances is constant.

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

Explanation:
1. Tangent to the parabola: \( y=mx+\frac2m \). The circle has centre \( (6,0) \) and \( r^2=32 \).
2. \( \frac{(6m+2/m)^2}{1+m^2}=32\Rightarrow(m^2-1)^2=0\Rightarrow m=\pm1 \). Tangents: \( y=x+2 \) and \( y=-x-2 \). Only \( y=x+2 \) is listed.

Teacher's Note:
1. Impose the tangency condition (distance from centre equals radius).
2. The resulting quartic is a perfect square.

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

Explanation:
1. A circle touching both axes has centre \( (\pm r,\pm r) \). On \( x+y=4 \): centre \( (r,r) \) with \( 2r=4 \), so \( r=2 \).
2. \( (x-2)^2+(y-2)^2=4\Rightarrow x^2+y^2-4x-4y+4=0 \).

Teacher's Note:
1. Touching both axes means \( |h|=|k|=r \).
2. Check each sign choice against the line.

Answer: (c) \( \frac13 \)

Explanation:
1. Let the radius be \( r \) and the centres \( C_A,C_B \) with \( C_AC_B=r \). The common points \( P,Q \) form equilateral triangles with the centres.
2. \( \angle PC_AQ=120^\circ \), so the smaller arc is \( \frac{120}{360}=\frac13 \) of the circumference.

Teacher's Note:
1. Triangle \( C_APC_B \) is equilateral with side \( r \).
2. The arc fraction equals the central angle divided by \( 360^\circ \).

Answer: (b) \( c<0 \)

Explanation:
1. Put \( y=0 \): \( x^2+2gx+c=0 \). The product of the roots is \( c \).
2. Roots on opposite sides of the origin have a negative product, so \( c<0 \).

Teacher's Note:
1. Opposite sides of the origin means roots of opposite sign.
2. \( c=0 \) would put one point at the origin.

Answer: (b) 6 units

Explanation:
1. Here \( a^2=4 \) (under \( x^2 \)) and \( b^2=9 \): the major axis is along the \( y \)-axis with semi-major axis 3.
2. Sum of focal distances \( =2\times3=6 \).

Teacher's Note:
1. Identify the larger denominator to find the major axis.
2. The sum of focal distances is the length of the major axis.

Answer: (d) No such point exists

Explanation:
1. The conjugate axis is the \( y \)-axis. Putting \( x=0 \): \( -9y^2=1 \), which has no real solution.
2. Hence there is no such point.

Teacher's Note:
1. The hyperbola meets its transverse axis but not its conjugate axis.
2. Set the other variable to zero and check for real roots.

Answer: (d) A hyperbola centred at \( (2,1) \) and of eccentricity \( \frac53 \)

Explanation:
1. \( \frac{(x-2)^2}9-\frac{(y-1)^2}{16}=\sec^2t-\tan^2t=1 \): a hyperbola with centre \( (2,1) \), \( a=3,b=4 \).
2. \( e=\sqrt{1+\frac{16}9}=\frac53 \).

Teacher's Note:
1. Use \( \sec^2t-\tan^2t=1 \) to eliminate \( t \).
2. For a hyperbola \( e=\sqrt{1+\frac{b^2}{a^2}} \).