JEE Mathematics Ellipse and Hyperbola MCQs Set 08

Answer: (c) \( 13 \)

Explanation:
1. Distance between the foci: \( \sqrt{12^2+5^2}=13=2ae \).
2. For a point of contact on the x-axis, \( 2a=PF_1+PF_2 \) is least when equal to the distance from \( F_1 \) to the image of \( F_2 \) in the x-axis, \( (14,-9) \): \( 2a=\sqrt{12^2+13^2}=\sqrt{313} \).
3. \( e=\frac{2ae}{2a}=\frac{13}{\sqrt{313}} \), so \( \lambda=13 \).

Teacher's Note:
1. A tangent line to an ellipse makes the sum of focal distances to its point of contact equal to \( 2a \), which is the shortest path from one focus to the line and then to the other. Reflect one focus in the line.

Answer: (d) (i), (iii) and (iv)

Explanation:
1. For \( y=2x+1 \) to touch the hyperbola: \( c^2=a^2m^2-b^2 \), so \( 1=4a^2-16 \) and \( a^2=\frac{17}{4} \).
2. (ii) \( (2a)^2=17=4^2+1^2 \), so the sides \( 2a,4,1 \) do form a right triangle.
3. (i) \( a^2=4.25 \): none of \( a^2=16+1 \), \( 16=a^2+1 \), \( 1=a^2+16 \) holds. (iii) \( a^2=4.25 \) versus \( 16+4=20 \), \( 16=a^2+4 \), \( 4=a^2+16 \): none. (iv) \( 4a^2=17 \) versus \( 64+1=65 \), \( 64=17+1 \), \( 1=17+64 \): none. So (i), (iii), (iv) cannot.

Teacher's Note:
1. Compute \( a^2 \) from the tangency condition and test the Pythagoras relation for each set. Only set (ii) satisfies it.

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

Explanation:
1. Distances of a vertex from the two foci are \( ae-a \) and \( ae+a \): \( a(e-1)=1 \) and \( a(e+1)=3 \), so \( a=1 \), \( e=2 \), \( b^2=a^2(e^2-1)=3 \).
2. If the transverse axis is the x-axis: \( x^2-\frac{y^2}{3}=1 \), i.e. \( 3x^2-y^2=3 \). If it is the y-axis: \( y^2-\frac{x^2}{3}=1 \), i.e. \( x^2-3y^2+3=0 \).
3. The other two have \( a^2=3,b^2=1 \) or \( a^2=6,b^2=2 \) and do not give the distances \( 1 \) and \( 3 \).

Teacher's Note:
1. The vertex-to-focus distances are \( a(e\pm1) \). The axes could be either coordinate axis, so check both orientations.

Answer: (d) \( x-3=0 \) and \( y-2=0 \)

Explanation:
1. \( xy-3y-2x=0 \) can be written as \( (x-3)(y-2)=6 \).
2. The asymptotes of \( (x-3)(y-2)=6 \) are obtained by setting the right side to zero.
3. So \( x-3=0 \) and \( y-2=0 \).

Teacher's Note:
1. A hyperbola \( xy+gx+fy+c=0 \) always factorises as \( (x+f)(y+g)=fg-c \), and the asymptotes are the two factor lines.

Answer: (a) P\( \to \)4; Q\( \to \)3; R\( \to \)1; S\( \to \)2

Explanation:
1. \( L=(0,b) \), \( M=(0,-b) \), \( N=(a,0) \). Half of the angle \( 60^\circ \) is \( 30^\circ \): \( \tan30^\circ=\frac ba \), so \( a=\sqrt3\,b \).
2. Area \( =\frac12(2b)(a)=ab=\sqrt3b^2=4\sqrt3 \), so \( b=2 \), \( a=2\sqrt3 \).
3. (P) conjugate axis \( 2b=4 \). (Q) \( e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{\frac43}=\frac2{\sqrt3} \). (R) \( 2ae=2\cdot2\sqrt3\cdot\frac2{\sqrt3}=8 \). (S) \( \frac{2b^2}{a}=\frac{8}{2\sqrt3}=\frac{4}{\sqrt3} \).

Teacher's Note:
1. The vertex \( N \) sees the conjugate axis under the angle \( 2\tan^{-1}\frac ba \). Everything else follows from \( a \) and \( b \).

Answer: (c) \( 2\sqrt7 \)

Explanation:
1. The line is \( y=-\frac34x+3\sqrt2 \), so \( m=-\frac34 \) and \( c=3\sqrt2 \). Tangency: \( c^2=a^2m^2+b^2 \), i.e. \( 18=\frac{9a^2}{16}+9 \), so \( a^2=16 \).
2. The major axis is along x: \( c_f^2=a^2-b^2=16-9=7 \).
3. Distance between the foci \( =2\sqrt7 \).

Teacher's Note:
1. Use the slope form of the tangency condition. Decide which of \( a^2 \) and \( b^2 \) is larger before computing the focal distance.

Answer: (d) \( 45\sqrt5 \)

Explanation:
1. The hyperbola is \( \frac{x^2}{9}-\frac{y^2}{36}=1 \). The chord of contact \( PQ \) from \( T(0,3) \) is \( -\frac{3y}{36}=1 \), i.e. \( y=-12 \).
2. Put \( y=-12 \): \( \frac{x^2}{9}=1+4=5 \), so \( P,Q=(\pm3\sqrt5,-12) \) and \( PQ=6\sqrt5 \).
3. Height of \( T \) above the chord: \( 3-(-12)=15 \). Area \( =\frac12\cdot6\sqrt5\cdot15=45\sqrt5 \).

Teacher's Note:
1. The chord of contact from an external point is \( T=0 \). Its distance from the point gives the height of the triangle.

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

Explanation:
1. For \( y=mx+c \) to touch both: \( c^2=16(1+m^2) \) and \( c^2=25m^2+4 \), so \( 9m^2=12 \) and \( m^2=\frac43 \). In the first quadrant \( m=-\frac{2}{\sqrt3} \), \( c^2=\frac{112}{3} \), \( c=\frac{4\sqrt7}{\sqrt3} \).
2. The line is \( y=-\frac{2}{\sqrt3}x+\frac{4\sqrt7}{\sqrt3} \), i.e. \( 2x+\sqrt3y=4\sqrt7 \).
3. Intercepts \( 2\sqrt7 \) and \( \frac{4\sqrt7}{\sqrt3} \): \( \lambda_1=\sqrt{28+\frac{112}{3}}=\frac{14}{\sqrt3} \). So (i) and (ii).

Teacher's Note:
1. For a common tangent in the first quadrant the slope must be negative. Check the intercept length using the Pythagoras theorem on the two intercepts.

Answer: (c) \( 2x-y-2=0 \)

Explanation:
1. \( e=2 \) gives \( b^2=3a^2 \): \( \frac{x^2}{a^2}-\frac{y^2}{3a^2}=1 \). Through \( (4,6) \): \( \frac{16}{a^2}-\frac{12}{a^2}=1 \), so \( a^2=4 \), \( b^2=12 \).
2. Tangent at \( (4,6) \): \( \frac{4x}{4}-\frac{6y}{12}=1 \), i.e. \( x-\frac y2=1 \).
3. So \( 2x-y-2=0 \).

Teacher's Note:
1. \( e^2=1+\frac{b^2}{a^2} \) gives \( b^2=3a^2 \). The point then fixes \( a^2 \).

Answer: (a) \( 25 \)

Explanation:
1. The lines \( 4x\mp3y=0 \) are the asymptotes. Take \( P=(3\sec\theta,4\tan\theta) \) with tangent \( \frac{x\sec\theta}{3}-\frac{y\tan\theta}{4}=1 \).
2. Intersecting with each asymptote gives \( Q=\left(\frac{3}{\sec\theta-\tan\theta},\frac{4}{\sec\theta-\tan\theta}\right) \) and \( R=\left(\frac{3}{\sec\theta+\tan\theta},-\frac{4}{\sec\theta+\tan\theta}\right) \).
3. \( CQ=\frac{5}{\sec\theta-\tan\theta} \), \( CR=\frac{5}{\sec\theta+\tan\theta} \), so \( CQ\cdot CR=\frac{25}{\sec^2\theta-\tan^2\theta}=25=a^2+b^2 \).

Teacher's Note:
1. The product of the distances from the centre to where a tangent meets the asymptotes is always \( a^2+b^2 \).

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

Explanation:
1. Take the rectangle \( 0\le x\le a \), \( 0\le y\le b \). The sides of length \( a \) are \( y=0,y=b \); the others are \( x=0,x=a \).
2. Condition: \( y^2+(b-y)^2=2\left[x^2+(a-x)^2\right] \), i.e. \( 2\left(y-\frac b2\right)^2+\frac{b^2}2=4\left(x-\frac a2\right)^2+a^2 \).
3. So \( 4\left(x-\frac a2\right)^2-2\left(y-\frac b2\right)^2=\frac{b^2}{2}-a^2 \): a hyperbola if \( b^2\neq2a^2 \) and a pair of lines if \( b^2=2a^2 \).

Teacher's Note:
1. The coefficients of the squared terms have opposite signs, so the locus cannot be a circle or an ellipse. The right-hand side decides between a hyperbola and a pair of lines.

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

Explanation:
1. Distance between the foci: \( \sqrt{19^2+5^2}=\sqrt{386}=2ae \). Distances of the origin from the foci: \( 13 \) and \( 25 \).
2. If the conic is an ellipse, \( 2a=13+25=38 \) and \( e=\frac{\sqrt{386}}{38} \). If it is a hyperbola, \( 2a=25-13=12 \) and \( e=\frac{\sqrt{386}}{12} \).
3. Both are possible, so (i) and (iv).

Teacher's Note:
1. A conic with given foci through a given point can be an ellipse (sum of focal distances) or a hyperbola (difference of focal distances). Both cases must be checked.

Answer: (c) \( 65 \)

Explanation:
1. Triangle \( OCF \) has a right angle at \( O \) with legs \( OF=6 \), \( OC=b \) and hypotenuse \( CF=a \). The inradius is \( 1 \).
2. For a right triangle \( r=\frac{b+6-a}{2}=1 \), so \( a=b+4 \). With \( a^2=b^2+36 \): \( 8b+16=36 \), so \( b=\frac52 \) and \( a=\frac{13}{2} \).
3. \( AB\cdot CD=2a\cdot2b=4\cdot\frac{13}{2}\cdot\frac52=65 \).

Teacher's Note:
1. Use the inradius formula \( r=\frac{\text{sum of legs}-\text{hypotenuse}}{2} \) for a right triangle. Note that the data give the diameter, so \( r=1 \).

Answer: (b) \( 2\sqrt5 \)

Explanation:
1. The centre is \( (0,3) \) and \( c=2 \). The given point lies on the line \( x=0 \), which is the minor axis, so it is an end of the minor axis and \( b=\sqrt5 \).
2. Check: \( a^2=b^2+c^2=9 \) and the sum of focal distances is \( 2\sqrt{4+5}=6=2a \).
3. The minor axis has length \( 2b=2\sqrt5 \).

Teacher's Note:
1. The perpendicular bisector of the foci is the minor axis, so a point on it at distance \( b \) from the centre is an end of the minor axis.

Answer: (c) \( (3,2) \) and \( \left(\frac95,\frac25\right) \)

Explanation:
1. The chord of contact from \( (2,1) \) is \( \frac{2x}{3}-\frac{y}{2}=1 \), i.e. \( 4x-3y=6 \), so \( y=\frac{4x-6}{3} \).
2. Substituting \( y=\frac{4x-6}{3} \) in \( \frac{x^2}{3}-\frac{y^2}{2}=1 \) and multiplying by \( 18 \): \( 6x^2-(4x-6)^2=18 \), i.e. \( 5x^2-24x+27=0 \), so \( x=3 \) or \( x=\frac95 \).
3. Corresponding \( y \): \( 2 \) and \( \frac25 \). The points are \( (3,2) \) and \( \left(\frac95,\frac25\right) \).

Teacher's Note:
1. The points of contact lie on the chord of contact \( T=0 \). Intersect it with the hyperbola and solve the quadratic.

Answer: (c) \( 4x-2y=1 \)

Explanation:
1. \( \frac ae=4 \) with \( e=\frac12 \) gives \( a=2 \) and \( b^2=a^2(1-e^2)=3 \): the ellipse is \( \frac{x^2}{4}+\frac{y^2}{3}=1 \). The point \( \left(1,\frac32\right) \) lies on it.
2. The normal at \( (x_1,y_1) \) is \( \frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2 \): \( 4x-\frac{3y}{3/2}=1 \).
3. So \( 4x-2y=1 \).

Teacher's Note:
1. The directrix gives \( \frac ae \), and with \( e \) known you get \( a \) and \( b^2 \). The normal formula for an ellipse is \( \frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2 \).

Answer: (a) \( 1 \)

Explanation:
1. In polar form a point at distance \( r \) in direction \( \varphi \) satisfies \( \frac{1}{r^2}=\frac{\cos^2\varphi}{a^2}-\frac{\sin^2\varphi}{b^2} \).
2. For the perpendicular direction \( \varphi+90^\circ \): \( \frac1{CQ^2}=\frac{\sin^2\varphi}{a^2}-\frac{\cos^2\varphi}{b^2} \).
3. Adding: \( \frac1{CP^2}+\frac1{CQ^2}=\frac1{a^2}-\frac1{b^2} \), so \( \lambda=1 \).

Teacher's Note:
1. Using polar coordinates about the centre makes the perpendicular-radii condition easy. The sum is independent of the direction, hence the constant.

Answer: (a) \( 3 \)

Explanation:
1. Let \( A=\left(t,\frac1t\right) \) and \( B=\left(\frac1t,t\right) \) (the intersections are symmetric about \( y=x \)).
2. \( AB^2=2\left(t-\frac1t\right)^2=14 \), so \( \left(t-\frac1t\right)^2=7 \).
3. \( r^2=t^2+\frac1{t^2}=7+2=9 \), so \( r=3 \).

Teacher's Note:
1. The curves are symmetric about \( y=x \), so the two first-quadrant intersection points are mirror images of each other.

Answer: (b) \( 2xy=\lambda(y^2-b^2) \)

Explanation:
1. \( m_1+m_2=\frac{2hk}{h^2-a^2} \) and \( m_1m_2=\frac{k^2-b^2}{h^2-a^2} \).
2. \( \cot\theta_1+\cot\theta_2=\frac{m_1+m_2}{m_1m_2}=\frac{2hk}{k^2-b^2}=\lambda \).
3. The locus is \( 2xy=\lambda(y^2-b^2) \).

Teacher's Note:
1. Use the sum and product of the roots of the quadratic in \( m \). The cotangents are reciprocals of the slopes, so their sum is \( \frac{m_1+m_2}{m_1m_2} \).

Answer: (b) \( \frac{63}{16} \)

Explanation:
1. The ellipse is \( \frac{x^2}{25}+\frac{y^2}{9}=1 \). A point \( (5\cos\theta,3\sin\theta) \) has squared distance from \( (3,0) \): \( d^2=(5\cos\theta-3)^2+9\sin^2\theta=16\cos^2\theta-30\cos\theta+18 \).
2. This quadratic in \( \cos\theta \) has its minimum at \( \cos\theta=\frac{15}{16} \) (within \( [-1,1] \)): \( d^2_{\min}=18-\frac{225}{16}=\frac{63}{16} \).
3. The largest inscribed circle has \( r^2=\frac{63}{16} \).

Teacher's Note:
1. The largest circle centred inside a convex curve reaches the nearest boundary point. Here the nearest point is not a vertex because the minimum of the quadratic lies inside \( [-1,1] \).

Answer: (d) \( 55\sqrt2 \)

Explanation:
1. Tangent \( y=mx+\frac1m \) to the parabola is tangent to \( \frac{x^2}{16}+\frac{y^2}{6}=1 \) if \( \frac{1}{m^2}=16m^2+6 \), i.e. \( 16m^4+6m^2-1=0 \), so \( m^2=\frac18 \).
2. Contact points on the parabola: \( \left(\frac1{m^2},\frac2m\right)=(8,\pm4\sqrt2) \). On the ellipse: \( \left(-\frac{16m}{c},\frac6c\right) \) with \( c=\frac1m \) gives \( \left(-2,\pm\frac{3}{\sqrt2}\right) \).
3. \( ABCD \) is a trapezium with parallel vertical sides \( 8\sqrt2 \) and \( 3\sqrt2 \) at distance \( 10 \): area \( =\frac{(8\sqrt2+3\sqrt2)}{2}\cdot10=55\sqrt2 \).

Teacher's Note:
1. Both tangent conditions in slope form give a quadratic in \( m^2 \). The four contact points are symmetric about the x-axis, which gives a trapezium.

Answer: (a) the mid-point of \( PQ \)

Explanation:
1. The normal at \( A \) is perpendicular to the tangent \( PA \), so \( \angle PAQ=90^\circ \). Similarly \( \angle PBQ=90^\circ \).
2. Hence \( P,A,Q,B \) lie on a circle with \( PQ \) as diameter.
3. This circle passes through \( Q,A,B \), so the circumcentre of triangle \( QAB \) is the mid-point of \( PQ \).

Teacher's Note:
1. When two normals at the points of contact meet at \( Q \), the quadrilateral \( PAQB \) has two right angles at \( A \) and \( B \) and is therefore cyclic.

Answer: (b) \( \frac{11}{2} \)

Explanation:
1. The hyperbola is \( \frac{x^2}{2}-\frac{y^2}{3}=1 \). A tangent \( y=mx+c \) has \( c^2=2m^2-3 \). Through \( (\alpha,2) \): \( c=2-m\alpha \).
2. \( (2-m\alpha)^2=2m^2-3 \) gives \( m^2(\alpha^2-2)-4\alpha m+7=0 \).
3. Product of slopes: \( m_1m_2=\frac{7}{\alpha^2-2}=2 \), so \( \alpha^2=2+\frac72=\frac{11}{2} \).

Teacher's Note:
1. Substitute the point into the tangent condition to get a quadratic in \( m \); the product of its roots is the product of the tangents of the angles.

Answer: (a) \( \frac{15}{4} \)

Explanation:
1. \( P=(5\cos\theta,3\sin\theta) \) and \( Q=\left(5\cos(\theta+\tfrac\pi6),3\sin(\theta+\tfrac\pi6)\right) \).
2. Area \( =\frac12|x_1y_2-x_2y_1|=\frac12\cdot15\left|\cos\theta\sin(\theta+\tfrac\pi6)-\cos(\theta+\tfrac\pi6)\sin\theta\right|=\frac{15}{2}\sin\frac\pi6=\frac{15}{4} \).

Teacher's Note:
1. The area of the triangle formed by the centre and two points with eccentric angles differing by \( \varphi \) is \( \frac12ab\sin\varphi \), independent of \( \theta \).