JEE Mathematics Ellipse and Hyperbola MCQs Set 09

Answer: (d) (i) and (ii) only

Explanation:
1. The foot of the perpendicular from \( (0,1) \) to \( x+y=3 \) is \( P(1,2) \). Points of the line at distance \( \frac{2\sqrt2}{3} \) from \( P \): \( \left(\frac53,\frac43\right) \) and \( \left(\frac13,\frac83\right) \).
2. A line \( x+y=3 \) touches \( \frac{x^2}{A}+\frac{y^2}{B}=1 \) when \( A+B=9 \), at \( \left(\frac A3,\frac B3\right) \). The point \( \left(\frac53,\frac43\right) \) gives \( A=5 \), \( B=4 \), a valid \( E_1 \) (\( A \gt B \)); \( \left(\frac13,\frac83\right) \) gives \( A=1,B=8 \), a valid \( E_2 \).
3. \( e_1^2=1-\frac45=\frac15 \), \( e_2^2=1-\frac18=\frac78 \). Then \( e_1^2+e_2^2=\frac{43}{40} \), \( e_1e_2=\sqrt{\frac{7}{40}}=\frac{\sqrt7}{2\sqrt{10}} \), \( |e_1^2-e_2^2|=\frac{27}{40} \). So (i) and (ii).

Teacher's Note:
1. The tangency condition \( A+B=n^2/(l^2) \) for a line \( x+y=3 \) fixes the sum of the squares of the semi-axes. The orientation of each ellipse decides which touch point belongs to it.

Answer: (a) \( 15 \)

Explanation:
1. The tangent at \( \theta \) is \( \frac{x\cos\theta}{9}+\frac{y\sin\theta}{6}=1 \) with intercepts \( \frac9{\cos\theta} \) and \( \frac6{\sin\theta} \).
2. The squared length is \( 81\sec^2\theta+36\csc^2\theta=81+36+81\tan^2\theta+36\cot^2\theta\ge117+2\cdot54=225 \).
3. The minimum length is \( 15=a+b \).

Teacher's Note:
1. The least length of the tangent segment between the axes of \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \) is always \( a+b \). AM-GM gives the minimum.

Answer: (c) \( \left(-1+\frac{2}{\sqrt6},\,2-\frac{1}{\sqrt6}\right) \)

Explanation:
1. The centre is \( (-1,2) \) and \( c=\sqrt{a^2+b^2}=\sqrt{\frac56} \).
2. The transverse axis \( x+2y-3=0 \) has direction \( (2,-1) \) with unit vector \( \frac{(2,-1)}{\sqrt5} \).
3. Foci: \( (-1,2)\pm\sqrt{\frac56}\cdot\frac{(2,-1)}{\sqrt5}=(-1,2)\pm\frac{(2,-1)}{\sqrt6} \). One focus is \( \left(-1+\frac2{\sqrt6},2-\frac1{\sqrt6}\right) \).

Teacher's Note:
1. The foci lie on the transverse axis, whose direction is \( (2,-1) \) for the line \( x+2y-3=0 \). Move a distance \( c \) from the centre in both directions.

Answer: (c) (i), (ii) and (iii) only

Explanation:
1. For \( P(x_1,y_1) \) the normal is \( y_1x+x_1y=2x_1y_1 \). It meets the x-axis at \( G(2x_1,0) \) and the y-axis at \( g(0,2y_1) \).
2. \( PG=\sqrt{x_1^2+y_1^2}=PC \) and \( Pg=\sqrt{x_1^2+y_1^2}=PC \), so \( PG=Pg=PC \): \( P \) is the mid-point of \( Gg \).
3. \( Gg=2PC\neq PC \), so (iv) is false.

Teacher's Note:
1. For a rectangular hyperbola, the point \( P \) is the mid-point of the part of its normal between the axes. Hence \( Gg=2PC \).

Answer: (d) \( \frac{2}{\sqrt3} \)

Explanation:
1. Conjugate axis \( 2b=\frac12(2ae)=ae \), so \( b=\frac{ae}{2} \) and \( b^2=\frac{a^2e^2}{4} \).
2. Also \( b^2=a^2(e^2-1) \). Equating: \( \frac{e^2}{4}=e^2-1 \), so \( e^2=\frac43 \).
3. \( e=\frac{2}{\sqrt3} \) (the latus rectum condition fixes \( a \) but not \( e \)).

Teacher's Note:
1. The condition on the conjugate axis alone determines \( e \). The latus rectum length only fixes the scale of the hyperbola.

Answer: (a) \( 0 \)

Explanation:
1. Completing squares: \( x^2-2x+1-(y^2-4y+4)=1 \) gives \( x^2-y^2-2x+4y-4=0 \), so the equation is \( (x-1)^2-(y-2)^2=1 \).
2. The centre is \( (1,2) \), which is the given point.
3. No real tangent can be drawn to a hyperbola from its centre.

Teacher's Note:
1. Completing the square shows that the given point is the centre of the hyperbola. Tangents from the centre would have to be asymptotes, which touch only at infinity.

Answer: (b) (i), (ii), (iii) and (iv)

Explanation:
1. The auxiliary circle of a hyperbola is the circle on its transverse axis as diameter, so here the semi-transverse axis must be \( 4 \).
2. (i) is \( \frac{x^2}{16}-\frac{y^2}{9}=1 \) and (ii) is \( \frac{y^2}{16}-\frac{x^2}{9}=1 \), both with \( a=4 \).
3. With \( u=\frac{x-y}{\sqrt2} \), \( v=\frac{x+y}{\sqrt2} \): (iii) is \( \frac{u^2}{16}-\frac{v^2}{9}=1 \) and (iv) is \( \frac{v^2}{16}-\frac{u^2}{9}=1 \). All have \( a=4 \) and the same centre.

Teacher's Note:
1. The same hyperbola can be turned or reflected, so the auxiliary circle stays the same as long as the centre and the semi-transverse axis are unchanged.

Answer: (d) \( \frac95 \)

Explanation:
1. Put \( u=\frac{x-2y+1}{\sqrt5} \) and \( v=\frac{2x+y+2}{\sqrt5} \) (perpendicular distances from two perpendicular lines). The equation becomes \( 20u^2+45v^2=25 \), i.e. \( \frac{u^2}{5/4}+\frac{v^2}{5/9}=1 \).
2. \( e^2=1-\frac{5/9}{5/4}=\frac59 \). The centre is where \( u=v=0 \): \( x-2y+1=0 \) and \( 2x+y+2=0 \) give \( (-1,0) \), at distance \( 1 \) from the origin.
3. \( 1=\lambda e \), so \( \lambda^2=\frac{1}{e^2}=\frac95 \).

Teacher's Note:
1. Rotated-axis problems become simple once you write each bracket as a distance from a line. The two lines \( x-2y+1=0 \) and \( 2x+y+2=0 \) are perpendicular.

Answer: (a) \( 9 \)

Explanation:
1. Let \( P=(4\cos\theta,3\sin\theta) \). The tangent \( \frac{x\cos\theta}{4}+\frac{y\sin\theta}{3}=1 \) gives \( CF=\frac{1}{\sqrt{\frac{\cos^2\theta}{16}+\frac{\sin^2\theta}{9}}} \).
2. The normal meets the x-axis at \( G=\left(\frac{(a^2-b^2)\cos\theta}{a},0\right)=\left(\frac{7\cos\theta}{4},0\right) \), so \( PG^2=\left(4\cos\theta-\tfrac{7\cos\theta}{4}\right)^2+9\sin^2\theta=\frac{81\cos^2\theta}{16}+9\sin^2\theta \).
3. \( CF^2\cdot PG^2=\frac{81\left(\frac{\cos^2\theta}{16}+\frac{\sin^2\theta}{9}\right)}{\frac{\cos^2\theta}{16}+\frac{\sin^2\theta}{9}}=81 \), so \( CF\cdot PG=9=b^2 \).

Teacher's Note:
1. For every ellipse \( CF\cdot PG=b^2 \). This is a classic property worth remembering along with \( CF\cdot CG'=a^2 \) for the corresponding quantities.

Answer: (d) \( \frac{13}{17} \)

Explanation:
1. \( F_1F_2 \) is a diameter of the circle, so \( \angle F_1PF_2=90^\circ \). For a point on the ellipse with this angle, area \( =b^2\tan\frac{90^\circ}{2}=b^2 \), so \( b^2=30 \).
2. \( a=\frac{17}{2} \), so \( e^2=1-\frac{b^2}{a^2}=1-\frac{120}{289}=\frac{169}{289} \).
3. \( e=\frac{13}{17} \).

Teacher's Note:
1. The area of triangle \( PF_1F_2 \) for a point \( P \) on an ellipse is \( b^2\tan\frac\varphi2 \) where \( \varphi=\angle F_1PF_2 \). A right angle gives area \( b^2 \).

Answer: (a) an ellipse whose eccentricity is \( \sqrt{\frac{2}{r+1}} \) when \( r \gt 1 \)

Explanation:
1. For \( r \gt 1 \): \( 1-r \lt 0 \), so the equation is \( \frac{y^2}{r+1}+\frac{x^2}{r-1}=1 \), an ellipse with major axis along y (\( r+1 \gt r-1 \)).
2. \( e^2=1-\frac{r-1}{r+1}=\frac{2}{r+1} \), so \( e=\sqrt{\frac{2}{r+1}} \).
3. For \( 0 \lt r \lt 1 \) it is a hyperbola with \( e^2=1+\frac{1-r}{1+r}=\frac{2}{1+r} \), i.e. \( e=\sqrt{\frac2{1+r}} \), which matches neither hyperbola option.

Teacher's Note:
1. Check the signs of both denominators to decide the type of conic. For an ellipse, use the larger denominator as \( a^2 \).

Answer: (c) \( \frac65 \)

Explanation:
1. Director circle radii: ellipse \( \sqrt{a^2+b^2}=2r \), hyperbola \( \sqrt{a^2-b^2}=r \).
2. So \( a^2+b^2=4(a^2-b^2) \), i.e. \( 3a^2=5b^2 \) and \( \frac{b^2}{a^2}=\frac35 \).
3. \( e_e^2=1-\frac35=\frac25 \) and \( e_h^2=1+\frac35=\frac85 \). The difference is \( \frac65 \).

Teacher's Note:
1. The director circle of an ellipse is \( x^2+y^2=a^2+b^2 \) and of a hyperbola \( x^2+y^2=a^2-b^2 \). Only the ratio \( b^2/a^2 \) is needed.

Answer: (a) \( 1 \)

Explanation:
1. The hyperbola is \( x^2-\frac{y^2}{4}=1 \) with asymptotes \( y=\pm2x \).
2. The point \( (1,2) \) satisfies \( y=2x \), so it lies on an asymptote (other than at the centre).
3. From a point on an asymptote only one real tangent can be drawn.

Teacher's Note:
1. From a point on an asymptote (not the centre) exactly one real tangent exists; from the centre none. Always test whether the point is on an asymptote.

Answer: (c) \( 2 \)

Explanation:
1. For \( P=(a\cos\theta,b\sin\theta) \): \( SS'=2ae \), \( PS=a-ae\cos\theta \), \( PS'=a+ae\cos\theta \). The incentre divides in the ratio of the sides, giving \( x=ae\cos\theta \) and \( y=\frac{eb\sin\theta}{1+e} \).
2. The locus is \( \frac{x^2}{(ae)^2}+\frac{y^2}{\left(\frac{eb}{1+e}\right)^2}=1 \) with \( e=e_1 \).
3. Its eccentricity: \( e_2^2=1-\frac{b^2}{a^2(1+e)^2}=1-\frac{1-e}{1+e}=\frac{2e}{1+e} \). Hence \( \left(1+\frac1e\right)e_2^2=\frac{1+e}{e}\cdot\frac{2e}{1+e}=2 \).

Teacher's Note:
1. Compute the incentre with weights equal to the opposite side lengths. Then compare \( ae \) with \( \frac{eb}{1+e} \) to see which is the semi-major axis.

Answer: (b) \( 2x-y+4+\sqrt{\frac32}=0 \)

Explanation:
1. \( a^2=\frac12 \), \( b^2=\frac13 \), so \( c^2=\frac56 \) and the distance of a directrix from the centre is \( \frac{a^2}{c}=\frac{1/2}{\sqrt{5/6}}=\frac12\sqrt{\frac65} \).
2. A directrix is parallel to the conjugate axis \( 2x-y+4=0 \) at this distance: \( \frac{2x-y+4}{\sqrt5}=\pm\frac12\sqrt{\frac65} \).
3. So \( 2x-y+4=\pm\frac{\sqrt6}{2}=\pm\sqrt{\frac32} \), i.e. \( 2x-y+4+\sqrt{\frac32}=0 \) is one directrix.

Teacher's Note:
1. Directrices are perpendicular to the transverse axis, so they are parallel to the conjugate axis. Their distance from the centre is \( \frac{a^2}{c}=\frac ae \).

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

Explanation:
1. The tangent at \( (x_1,y_1) \) to \( x^2+y^2=2 \) is \( xx_1+yy_1=2 \).
2. At \( (-1,1) \): \( -x+y=2 \), i.e. \( y=x+2 \) (slope \( 1 \), intercept \( 2=a\sqrt{1+m^2} \)).

Teacher's Note:
1. For a circle \( x^2+y^2=a^2 \) the tangent at \( (x_1,y_1) \) is \( xx_1+yy_1=a^2 \). In slope form \( y=mx\pm a\sqrt{1+m^2} \).

Answer: (b) \( \frac{5\sqrt5}{2} \)

Explanation:
1. The ellipse is \( \frac{x^2}{4}+\frac{y^2}{3}=1 \). The normal at \( (x_1,y_1) \) has slope \( \frac{4y_1}{3x_1} \). Setting this equal to \( -2 \): \( y_1=-\frac{3x_1}{2} \).
2. On the ellipse: \( \frac{x_1^2}{4}+\frac{3x_1^2}{4}=1 \), so \( x_1=\pm1 \): \( P=(1,-\tfrac32) \) or \( \left(-1,\tfrac32\right) \).
3. The tangent \( \frac{xx_1}{4}+\frac{yy_1}{3}=1 \) passes through \( (4,4) \) only for \( P=\left(-1,\frac32\right) \) (value \( -1+2=1 \)). \( PQ=\sqrt{25+\frac{25}{4}}=\frac{5\sqrt5}{2} \).

Teacher's Note:
1. Two candidate points arise from the normal condition. The tangent condition then picks the correct one.

Answer: (d) (ii) and (iii) only

Explanation:
1. The largest rectangle in \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \) has vertices \( \left(\pm\frac a{\sqrt2},\pm\frac b{\sqrt2}\right) \) and area \( 2ab \). The largest ellipse inscribed in a rectangle with half-sides \( p,q \) is \( \frac{x^2}{p^2}+\frac{y^2}{q^2}=1 \).
2. So \( a_{n+1}=\frac{a_n}{\sqrt2} \), \( b_{n+1}=\frac{b_n}{\sqrt2} \): all \( E_n \) are similar and have the same eccentricity (so (i) is false). \( a_9=\frac{3}{16} \), \( b_9=\frac{1}{8} \).
3. (ii) Latus rectum of \( E_9 \): \( \frac{2b^2}{a}=\frac{2/64}{3/16}=\frac16 \) (true). (iii) Area of \( R_n=2a_nb_n=\frac{12}{2^{n-1}} \), sum \( \lt 24 \) (true). (iv) \( c=\sqrt{\frac{9}{256}-\frac{4}{256}}=\frac{\sqrt5}{16} \), not \( \frac{\sqrt5}{32} \) (false).

Teacher's Note:
1. Each step shrinks both semi-axes by the same factor \( \sqrt2 \), so the shape (and the eccentricity) never changes. The areas of \( R_n \) form a geometric series with ratio \( \frac12 \).

Answer: (c) \( \frac{1}{2x^2}+\frac{1}{4y^2}=1 \)

Explanation:
1. The ellipse is \( \frac{x^2}{2}+y^2=1 \). The tangent at \( (\sqrt2\cos\theta,\sin\theta) \) is \( \frac{x\cos\theta}{\sqrt2}+y\sin\theta=1 \) with intercepts \( \frac{\sqrt2}{\cos\theta} \) and \( \frac1{\sin\theta} \).
2. Mid-point \( (h,k)=\left(\frac{1}{\sqrt2\cos\theta},\frac{1}{2\sin\theta}\right) \), so \( \cos\theta=\frac{1}{\sqrt2h} \) and \( \sin\theta=\frac1{2k} \).
3. Using \( \cos^2\theta+\sin^2\theta=1 \): \( \frac1{2h^2}+\frac1{4k^2}=1 \).

Teacher's Note:
1. Parametrise the tangent, find the intercepts, take half of each and eliminate \( \theta \). The four vertices are excluded because there one intercept is infinite.

Answer: (a) (ii), (iii) and (iv)

Explanation:
1. The origin gives \( 2x^2-3y^2-6=-6 \lt 0 \), so the point must satisfy \( 2x^2-3y^2-6 \lt 0 \).
2. \( 18\sin^2\alpha-12\cos^2\alpha-6 \lt 0 \) gives \( 30\sin^2\alpha \lt 18 \), i.e. \( \sin^2\alpha \lt \frac35 \).
3. \( \frac25 \lt \frac35 \), \( \frac15 \lt \frac35 \) and \( \frac{4}{225} \lt \frac35 \) qualify; \( \frac45 \gt \frac35 \) does not. So (ii), (iii) and (iv).

Teacher's Note:
1. Two points are on the same side of a curve when \( S(x,y) \) has the same sign for both. Evaluate \( S \) at the origin first.

Answer: (c) \( (2\sqrt2,3\sqrt3) \)

Explanation:
1. Let \( a^2=A \); \( b^2=4-A \). Passing through \( P \): \( \frac2A-\frac{3}{4-A}=1 \), so \( A^2-9A+8=0 \) and \( A=1 \) (as \( A \lt 4 \)). The hyperbola is \( x^2-\frac{y^2}{3}=1 \).
2. Tangent at \( P \): \( \sqrt2x-\frac{\sqrt3y}{3}=1 \).
3. At \( (2\sqrt2,3\sqrt3) \): \( 4-3=1 \). This point lies on the tangent.

Teacher's Note:
1. Use \( a^2+b^2=c^2=4 \) and the point on the hyperbola to find \( a^2 \). Substitute the options into the tangent equation.

Answer: (b) (i) and (iii) only

Explanation:
1. Slope of \( AB=\frac{b(\sin\beta-\sin\alpha)}{a(\cos\beta-\cos\alpha)}=-\frac ba\cot\frac{\alpha+\beta}{2} \).
2. Slope of the tangent at \( \theta \) is \( -\frac ba\cot\theta \).
3. So \( \cot\theta=\cot\frac{\alpha+\beta}2 \), giving \( \theta=\frac{\alpha+\beta}{2}+n\pi \). The values (i) and (iii) are of this form.

Teacher's Note:
1. A chord and the tangent at the point of the ellipse with the mean eccentric angle are parallel. Adding \( \pi \) gives the diametrically opposite point, whose tangent is also parallel.

Answer: (b) (i), (ii) and (iv)

Explanation:
1. The tangent at \( P \) is \( xx_1-yy_1=1 \), so \( M=\left(\frac1{x_1},0\right) \). The normal at \( P \) passes through \( N \) and has direction \( (x_1,-y_1) \), so \( N=(2x_1,0) \).
2. Centroid: \( l=\frac{x_1+\frac1{x_1}+2x_1}{3}=x_1+\frac{1}{3x_1} \) and \( m=\frac{y_1}{3} \).
3. \( \frac{dl}{dx_1}=1-\frac{1}{3x_1^2} \), \( m=\frac{\sqrt{x_1^2-1}}{3} \) gives \( \frac{dm}{dx_1}=\frac{x_1}{3\sqrt{x_1^2-1}} \) and \( \frac{dm}{dy_1}=\frac13 \). So (i), (ii), (iv).

Teacher's Note:
1. A circle touching a hyperbola at \( P \) has its centre on the normal at \( P \). Use this to find \( N \) before computing the centroid.