JEE Mathematics Ellipse and Hyperbola MCQs Set 10

Answer: (a) \( \left(\frac{25}{4},\frac{16}{3}\right) \)

Explanation:
1. The line \( PQ \) is the chord of contact of the required point \( (x_1,y_1) \): \( \frac{x_1x}{25}+\frac{y_1y}{16}=1 \).
2. Comparing with \( \frac{3x}{12}+\frac{4y}{12}=1 \): \( \frac{x_1}{25}=\frac14 \) and \( \frac{y_1}{16}=\frac13 \).
3. So \( (x_1,y_1)=\left(\frac{25}{4},\frac{16}{3}\right) \).

Teacher's Note:
1. The chord joining the contact points is the polar of the pole. Match coefficients.

Answer: (c) \( 16 \)

Explanation:
1. The hyperbola is \( \frac{x^2}{144/25}-\frac{y^2}{81/25}=1 \), so \( c^2=\frac{144+81}{25}=9 \) and the foci are \( (\pm3,0) \).
2. For the ellipse \( 25-b^2=9 \), so \( b^2=16 \).

Teacher's Note:
1. Rewrite the hyperbola in standard form first by multiplying through by \( 25 \).

Answer: (b) \( x+y=\pm3\sqrt3 \)

Explanation:
1. The hyperbola is \( \frac{x^2}{36}-\frac{y^2}{9}=1 \). A line perpendicular to \( x-y+4=0 \) has slope \( m=-1 \).
2. \( c^2=a^2m^2-b^2=36-9=27 \), so \( c=\pm3\sqrt3 \).
3. The tangents are \( y=-x\pm3\sqrt3 \), i.e. \( x+y=\pm3\sqrt3 \).

Teacher's Note:
1. Perpendicular slope is the negative reciprocal. Then use the hyperbola tangency condition.

Answer: (d) \( 3x+5y=25 \)

Explanation:
1. \( e=\frac35 \) and the end of the latus rectum is \( \left(ae,\frac{b^2}{a}\right)=\left(3,\frac{16}{5}\right) \).
2. Tangent: \( \frac{3x}{25}+\frac{(16/5)y}{16}=1 \), i.e. \( \frac{3x}{25}+\frac y5=1 \), so \( 3x+5y=25 \).
3. It meets the x-axis at \( \left(\frac{25}{3},0\right)=\left(\frac ae,0\right) \), the point where the directrix meets the major axis.

Teacher's Note:
1. Use the tangent formula \( \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1 \) at \( (ae,b^2/a) \).

Answer: (a) \( 7x^2+12xy-2y^2-2x+14y-22=0 \)

Explanation:
1. For a point \( P(x,y) \): \( PS^2=e^2\,PM^2 \), so \( (x-1)^2+(y-2)^2=3\cdot\frac{(2x+y-1)^2}{5} \).
2. \( 5\left(x^2+y^2-2x-4y+5\right)=3\left(4x^2+y^2+4xy-4x-2y+1\right) \).
3. Simplifying: \( 7x^2+12xy-2y^2-2x+14y-22=0 \).

Teacher's Note:
1. Use the focus-directrix definition \( PS=e\cdot PM \) and square it to remove the root.

Answer: (c) \( y=\pm x\pm\sqrt{10} \) (four lines)

Explanation:
1. Circle: \( c^2=5(1+m^2) \). Ellipse \( \frac{x^2}{9}+y^2=1 \): \( c^2=9m^2+1 \).
2. Equating: \( 5+5m^2=9m^2+1 \), so \( m^2=1 \) and \( c^2=10 \).
3. The tangents are \( y=\pm x\pm\sqrt{10} \).

Teacher's Note:
1. A line \( y=mx+c \) touches both conics only if both tangency conditions hold. Equate the two values of \( c^2 \).

Answer: (b) \( x^2+4y^2=20 \) or \( x^2+64y^2=80 \)

Explanation:
1. Let it be \( \frac{x^2}{A}+\frac{y^2}{B}=1 \). Through \( (4,-1) \): \( \frac{16}{A}+\frac1B=1 \).
2. Tangency to \( x+4y=10 \): \( A\cdot1^2+B\cdot4^2=10^2 \), so \( A=100-16B \).
3. Substituting: \( 4B^2-25B+25=0 \), so \( B=5 \) (then \( A=20 \)) or \( B=\frac54 \) (then \( A=80 \)).
4. The ellipses are \( x^2+4y^2=20 \) and \( x^2+64y^2=80 \).

Teacher's Note:
1. The condition for \( lx+my=n \) to touch \( \frac{x^2}{A}+\frac{y^2}{B}=1 \) is \( Al^2+Bm^2=n^2 \). Two unknowns, two equations, so two solutions are possible.

Answer: (d) \( \frac{\sqrt{13}}{3} \)

Explanation:
1. Let it be \( \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \). The point \( (3,0) \) gives \( a^2=9 \).
2. The point \( (3\sqrt2,2) \): \( \frac{18}{9}-\frac{4}{b^2}=1 \), so \( b^2=4 \).
3. \( e^2=1+\frac49=\frac{13}{9} \), so \( e=\frac{\sqrt{13}}{3} \).

Teacher's Note:
1. A point on the axis fixes \( a^2 \). The second point then fixes \( b^2 \). Do not forget the square root in \( e \).

Answer: (b) touch each other at \( (5,4) \)

Explanation:
1. The line is \( y=-\frac45x+8 \). The condition for tangency is \( c^2=a^2m^2+b^2 \): \( 64 \) and \( 50\cdot\frac{16}{25}+32=64 \). They are equal, so the line touches.
2. The point of contact is \( \left(-\frac{a^2m}{c},\frac{b^2}{c}\right)=\left(\frac{50\cdot4/5}{8},\frac{32}{8}\right)=(5,4) \).

Teacher's Note:
1. Compare \( c^2 \) with \( a^2m^2+b^2 \). If equal, the line is a tangent and the contact point comes from the standard formula.

Answer: (a) \( (x^2+y^2)^2=16x^2-9y^2 \)

Explanation:
1. The chord of the circle with mid-point \( (h,k) \) is \( hx+ky=h^2+k^2 \).
2. The hyperbola is \( \frac{x^2}{16}-\frac{y^2}{9}=1 \). A line \( lx+my=n \) touches it if \( 16l^2-9m^2=n^2 \).
3. So \( 16h^2-9k^2=(h^2+k^2)^2 \), i.e. \( (x^2+y^2)^2=16x^2-9y^2 \).

Teacher's Note:
1. Combine \( T=S_1 \) for the circle with the tangency condition \( a^2l^2-b^2m^2=n^2 \) for the hyperbola.

Answer: (a) \( (2,3) \) and \( (6,7) \)

Explanation:
1. The major axis passes through the focus perpendicular to the directrix, so it has direction \( (1,1) \): points \( \left(3+\frac{s}{\sqrt2},4+\frac{s}{\sqrt2}\right) \).
2. Vertices satisfy \( s^2=e^2\cdot d^2 \), where \( d=\frac{|x+y-1|}{\sqrt2}=\frac{|6+\sqrt2 s|}{\sqrt2} \). Solving gives \( s=-\sqrt2 \) and \( s=3\sqrt2 \).
3. The points are \( (2,3) \) and \( (6,7) \). Check: at \( (2,3) \) the focal distance is \( \sqrt2 \) and the directrix distance is \( 2\sqrt2 \), a ratio of \( \frac12 \).

Teacher's Note:
1. The axis is the normal to the directrix through the focus. A vertex divides the focus-directrix segment in the ratio \( e:1 \) (internally or externally).

Answer: (a) \( (3,0) \)

Explanation:
1. Let \( P=(5\cos\theta,4\sin\theta) \). The tangent \( \frac{x\cos\theta}{5}+\frac{y\sin\theta}{4}=1 \) meets \( x=\frac{25}{3} \) at \( Q=\left(\frac{25}{3},\frac{4(1-\frac53\cos\theta)}{\sin\theta}\right) \).
2. The circle is \( (x-5\cos\theta)\left(x-\frac{25}{3}\right)+(y-4\sin\theta)(y-y_Q)=0 \). Put \( y=0 \): \( x^2-\left(5\cos\theta+\frac{25}{3}\right)x+15\cos\theta+16=0 \), which factors as \( (x-3)\left(x-5\cos\theta-\frac{16}{3}\right)=0 \).
3. So \( x=3 \) for every \( \theta \): the circle passes through \( (3,0) \), the focus corresponding to the directrix \( x=\frac{25}{3} \).

Teacher's Note:
1. The tangent at \( P \) meets the directrix at \( Q \) with \( SP\perp SQ \), where \( S \) is the matching focus. So the circle on \( PQ \) passes through \( S \).

Answer: (d) \( e=\frac53 \); foci \( (4,2) \) and \( (-6,2) \)

Explanation:
1. Complete squares: \( 16(x+1)^2-9(y-2)^2=164+16-36=144 \), so \( \frac{(x+1)^2}{9}-\frac{(y-2)^2}{16}=1 \).
2. Centre \( (-1,2) \), \( a=3 \), \( b=4 \), \( e=\sqrt{1+\frac{16}{9}}=\frac53 \).
3. Foci: \( (-1\pm ae,2)=(-1\pm5,2) \), i.e. \( (4,2) \) and \( (-6,2) \).

Teacher's Note:
1. Complete the square in \( x \) and \( y \) to reach the standard form. The transverse axis is along the positive term.

Answer: (d) \( \frac45 \lt \alpha \lt \frac{4}{\sqrt{17}} \)

Explanation:
1. The line meets the axes at \( (4,0) \) and \( (0,3) \), which are on the ellipse, so it cuts off the smaller region on the side away from the origin: \( 3x+4y \gt 12 \).
2. \( 3\alpha+12\alpha \gt 12 \) gives \( \alpha \gt \frac45 \).
3. Inside the ellipse: \( 9\alpha^2+144\alpha^2 \lt 144 \), so \( \alpha^2 \lt \frac{16}{17} \) and \( \alpha \lt \frac{4}{\sqrt{17}} \).

Teacher's Note:
1. The origin is on the larger side of the chord. The smaller region is where the line expression exceeds \( 12 \) and the ellipse expression is below \( 144 \).

Answer: (b) \( \frac{12}{5} \lt \alpha \lt \frac{16}{5} \)

Explanation:
1. Inside means \( \frac{\left(7-\frac54\alpha\right)^2}{25}+\frac{\alpha^2}{16} \lt 1 \).
2. Expanding: \( \frac{49}{25}-\frac{7\alpha}{10}+\frac{\alpha^2}{16}+\frac{\alpha^2}{16} \lt 1 \), which becomes \( 25\alpha^2-140\alpha+192 \lt 0 \).
3. The roots are \( \alpha=\frac{12}{5} \) and \( \frac{16}{5} \), so the point is strictly inside for \( \frac{12}{5} \lt \alpha \lt \frac{16}{5} \).

Teacher's Note:
1. A point is inside the ellipse when \( S_1 \lt 0 \). The boundary values are on the ellipse, so they are excluded.

Answer: (c) \( 4r^2=(2+\sqrt2)(a^2p^2+b^2q^2) \)

Explanation:
1. The chord through the points with eccentric angles \( \alpha,\beta \) is \( \frac xa\cos\frac{\alpha+\beta}{2}+\frac yb\sin\frac{\alpha+\beta}{2}=\cos\frac{\alpha-\beta}{2} \).
2. Write the given line as \( (pa)\frac xa+(qb)\frac yb=r \). Comparing: \( \cos\frac{\alpha-\beta}{2}=\frac{r}{\sqrt{a^2p^2+b^2q^2}} \).
3. With \( \alpha-\beta=\frac\pi4 \): \( \cos^2\frac\pi8=\frac{2+\sqrt2}{4} \), so \( r^2=\frac{(2+\sqrt2)}{4}(a^2p^2+b^2q^2) \).

Teacher's Note:
1. Normalise the coefficients so the line matches the eccentric-angle chord. Then use \( \cos^2\frac\pi8=\frac{1+\cos\frac\pi4}{2} \).

Answer: (a) \( 2 \)

Explanation:
1. Let the ellipse be \( \frac{x^2}{a_1^2}+\frac{y^2}{b^2}=1 \) and the hyperbola \( \frac{x^2}{a_2^2}-\frac{y^2}{b^2}=1 \) with common focal distance \( c \).
2. \( c^2=a_1^2-b^2=a_2^2+b^2 \), so \( a_1^2+a_2^2=2c^2 \).
3. \( \frac1{e_1^2}+\frac1{e_2^2}=\frac{a_1^2+a_2^2}{c^2}=2 \).

Teacher's Note:
1. Use \( e_1=c/a_1 \), \( e_2=c/a_2 \), and the common \( b \) to link the two relations.

Answer: (d) \( -\frac{16}{9} \) and \( -\frac{20}{9} \)

Explanation:
1. Tangent through \( (4,2) \): \( y-2=m(x-4) \), so \( c=2-4m \). Tangency: \( (2-4m)^2=25m^2-16 \).
2. \( 9m^2+16m-20=0 \).
3. Sum of roots \( =-\frac{16}{9} \), product \( =-\frac{20}{9} \).

Teacher's Note:
1. Impose \( c^2=a^2m^2-b^2 \) with \( c=y_1-mx_1 \). The slopes are the two roots of the resulting quadratic.

Answer: (c) \( 30^\circ \)

Explanation:
1. Tangent: \( \frac{x\sec\phi}{3}-\frac{y\tan\phi}{4}=1 \), with slope \( \frac{4\sec\phi}{3\tan\phi}=\frac{4}{3\sin\phi} \).
2. The given line has slope \( -\frac38 \), so the tangent has slope \( \frac83 \).
3. \( \frac{4}{3\sin\phi}=\frac83 \) gives \( \sin\phi=\frac12 \), so \( \phi=30^\circ \).

Teacher's Note:
1. Differentiate or read the slope from the tangent form. The perpendicular condition is \( m_1m_2=-1 \).

Answer: (c) \( \frac12 \)

Explanation:
1. Foci distance \( 2ae=10 \), so \( ae=5 \). Focus to directrix distance \( \frac ae-ae=15 \), so \( \frac ae=20 \).
2. Multiplying: \( a^2=100 \), and dividing: \( e^2=\frac{ae}{a/e}=\frac{5}{20}=\frac14 \).
3. So \( e=\frac12 \).

Teacher's Note:
1. The directrix is at distance \( a/e \) from the centre and the focus at \( ae \). Use their sum and difference to get \( e^2 \).

Answer: (c) \( \pm6 \)

Explanation:
1. The hyperbola is \( \frac{x^2}{5}-\frac{y^2}{9}=1 \) with \( a^2=5 \), \( b^2=9 \).
2. Tangency: \( c^2=a^2m^2-b^2=5\cdot9-9=36 \).
3. So \( \lambda=\pm6 \).

Teacher's Note:
1. For a hyperbola the tangency condition is \( c^2=a^2m^2-b^2 \).

Answer: (c) \( \frac{\sqrt7}{4} \)

Explanation:
1. On the x-axis \( 2x=6 \) gives \( (3,0) \). On the y-axis \( 5y=20 \) gives \( (0,4) \).
2. So the semi-axes are \( 3 \) (along \( x \)) and \( 4 \) (along \( y \)); the major axis is the \( y \) axis with \( a=4 \), \( b=3 \).
3. \( e=\sqrt{1-\frac{9}{16}}=\frac{\sqrt7}{4} \).

Teacher's Note:
1. The intercept points are the ends of the axes. The larger semi-axis is the semi-major one.

Answer: (c) \( 8 \)

Explanation:
1. Tangents from \( (3,2) \): \( x=3 \) (tangent at the vertex) and \( 5x-12y+9=0 \) (from \( m=\frac5{12} \)).
2. Chord of contact: \( 3x-18y=9 \), i.e. \( x-6y=3 \). It meets the hyperbola at \( (3,0) \) and \( \left(-5,-\frac43\right) \).
3. The triangle has vertices \( (3,2) \), \( (3,0) \), \( \left(-5,-\frac43\right) \): area \( =\frac12\cdot2\cdot8=8 \) (base \( 2 \) on \( x=3 \), height \( 8 \)).

Teacher's Note:
1. Find the contact points from the polar, then take the area of the triangle with the external point.

Answer: (d) \( (4,1) \) and \( (-6,1) \)

Explanation:
1. \( 9(x+1)^2-16(y-1)^2=151+9-16=144 \), so \( \frac{(x+1)^2}{16}-\frac{(y-1)^2}{9}=1 \).
2. Centre \( (-1,1) \), \( a=4 \), \( b=3 \), \( c=5 \), and the foci are \( (-1\pm5,1) \).

Teacher's Note:
1. Completing the square gives the centre. The transverse axis is horizontal, so the foci are offset from the centre in \( x \).

Answer: (b) \( 48x+25y=169 \)

Explanation:
1. Use \( T=S_1 \): \( \frac{3x}{25}+\frac{y}{16}-1=\frac{9}{25}+\frac{1}{16}-1 \).
2. Multiplying by \( 400 \): \( 48x+25y=144+25=169 \).

Teacher's Note:
1. The chord with mid-point \( (h,k) \) is \( T=S_1 \).

Answer: (b) \( 12x+5y=48 \)

Explanation:
1. The ellipse is \( \frac{x^2}{25}+\frac{y^2}{9}=1 \) with foci \( (\pm4,0) \), so the ray starts at a focus.
2. \( P=\left(3,\frac{12}{5}\right) \). A ray through one focus reflects through the other focus \( (4,0) \).
3. Line through \( (4,0) \) and \( \left(3,\frac{12}{5}\right) \): slope \( -\frac{12}{5} \), so \( y=-\frac{12}{5}(x-4) \), i.e. \( 12x+5y=48 \).

Teacher's Note:
1. The reflection property: rays through one focus reflect through the other focus.

Answer: (d) lies in the region containing the focus of the right branch (inside)

Explanation:
1. \( S_1=9(2)^2-5^2-1=10 \gt 0 \).
2. For a hyperbola \( S_1 \gt 0 \) means the point is in the region enclosed by a branch (containing a focus). Since \( x \gt 0 \), it is the right branch.
3. Check: at \( x=2 \) the branch has \( y=\pm\sqrt{35}\approx\pm5.9 \), and \( 5 \lt 5.9 \), so the point is within the branch.

Teacher's Note:
1. Substitute into \( S_1 \). For a hyperbola a positive value places the point inside a branch, a negative value between the branches.

Answer: (b) \( 4x+5y=40 \) and \( 4x-35y=200 \)

Explanation:
1. Let \( y+4=m(x-15) \), i.e. \( y=mx-(15m+4) \). Tangency: \( (15m+4)^2=50m^2+32 \).
2. \( 175m^2+120m-16=0 \) gives \( m=-\frac45 \) or \( m=\frac{4}{35} \).
3. \( m=-\frac45 \): \( 4x+5y=40 \). \( m=\frac4{35} \): \( 4x-35y=200 \).

Teacher's Note:
1. Use the point-slope form and the condition \( c^2=a^2m^2+b^2 \). The quadratic in \( m \) gives both tangents.

Answer: (a) \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{y}{b} \)

Explanation:
1. The chord with mid-point \( (h,k) \) is \( T=S_1 \): \( \frac{hx}{a^2}+\frac{ky}{b^2}-1=\frac{h^2}{a^2}+\frac{k^2}{b^2}-1 \).
2. It passes through \( (0,b) \): \( \frac{k}{b}=\frac{h^2}{a^2}+\frac{k^2}{b^2} \).
3. Replacing \( (h,k) \) by \( (x,y) \) gives \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac yb \).

Teacher's Note:
1. Use \( T=S_1 \) for the chord with a given mid-point and force it through the fixed point.

Answer: (d) \( \frac{2\sqrt2}{3} \)

Explanation:
1. \( x^2\left(\frac19-\frac1{25}\right)=1 \) gives \( x^2=\frac{225}{16} \), so \( P,Q=\pm\left(\frac{15}{4},\frac{15}{4}\right) \) and \( PQ=\frac{15}{\sqrt2} \).
2. So \( 2a=\frac{15}{\sqrt2} \) and \( 2b=\frac{5}{\sqrt2} \), hence \( \frac ba=\frac13 \).
3. \( e=\sqrt{1-\frac19}=\frac{2\sqrt2}{3} \).

Teacher's Note:
1. The length \( PQ \) is the major axis. Only the ratio \( b/a \) is needed for the eccentricity.

Answer: (a) \( \pm2 \)

Explanation:
1. The line is \( y=-\sqrt2x-\frac{n}{\sqrt2} \), so \( m^2=2 \) and \( c^2=\frac{n^2}{2} \).
2. Tangency: \( c^2=a^2m^2-b^2=18-16=2 \).
3. So \( n^2=4 \) and \( n=\pm2 \).

Teacher's Note:
1. Put the line in the form \( y=mx+c \) and use \( c^2=a^2m^2-b^2 \).

Answer: (b) \( \frac{x^2}{4}-\frac{y^2}{12}=1 \)

Explanation:
1. The ellipse has \( c=\sqrt{25-9}=4 \), so the foci are \( (\pm4,0) \).
2. For the hyperbola \( ae=4 \) and \( e=2 \) give \( a=2 \), \( a^2=4 \) and \( b^2=a^2(e^2-1)=12 \).

Teacher's Note:
1. Confocal conics share \( c \). For the hyperbola \( b^2=c^2-a^2 \).