JEE Mathematics Ellipse MCQs Set 09

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 \( \frac{x\cos\theta}{\sqrt2}+y\sin\theta=1 \).
2. Its intercepts are \( \frac{\sqrt2}{\cos\theta} \) and \( \frac1{\sin\theta} \), so the mid-point is \( x=\frac{1}{\sqrt2\cos\theta} \), \( y=\frac1{2\sin\theta} \).
3. Using \( \cos^2\theta+\sin^2\theta=1 \): \( \frac{1}{2x^2}+\frac{1}{4y^2}=1 \).

Teacher's Note:
1. Write the tangent in parametric form, find the axis intercepts and eliminate the parameter.

Answer: (d) \( (4\sqrt3,2\sqrt2) \)

Explanation:
1. \( 2ae=2b \) gives \( b=ae \), so \( b^2=a^2e^2=a^2(1-e^2) \) and \( e^2=\frac12 \), \( b^2=\frac{a^2}{2} \).
2. Latus rectum \( \frac{2b^2}{a}=a=8 \), so \( a^2=64 \), \( b^2=32 \) and the ellipse is \( \frac{x^2}{64}+\frac{y^2}{32}=1 \).
3. Check \( (4\sqrt3,2\sqrt2) \): \( \frac{48}{64}+\frac{8}{32}=1 \). The other points do not satisfy the equation.

Teacher's Note:
1. Convert each condition into a relation between a, b and e, then solve for a and b.

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

Explanation:
1. Directrix \( \frac{a}{e}=4 \) gives \( a=2 \); \( b^2=a^2(1-e^2)=3 \). The ellipse is \( \frac{x^2}{4}+\frac{y^2}{3}=1 \) and \( \left(1,\frac32\right) \) lies on it.
2. The normal at \( (x_1,y_1) \) is \( \frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2 \), i.e. \( 4x-2y=1 \).

Teacher's Note:
1. Find a and b from the eccentricity and directrix, then use the standard normal equation.

Answer: (c) \( \frac92 \)

Explanation:
1. At a common point \( (x_0,y_0) \) the slopes are \( \frac{3}{y_0} \) (parabola) and \( -\frac{9x_0}{by_0} \) (ellipse).
2. Perpendicular curves need the product \( =-1 \): \( -\frac{27x_0}{by_0^2}=-1 \). With \( y_0^2=6x_0 \) this gives \( \frac{27}{6b}=1 \), so \( b=\frac92 \).

Teacher's Note:
1. Orthogonal curves have tangent slopes whose product is -1 at the point of intersection.

Answer: (c) \( (x^2+y^2)^2=6x^2+2y^2 \)

Explanation:
1. The ellipse is \( \frac{x^2}{6}+\frac{y^2}{2}=1 \), so \( a^2=6 \), \( b^2=2 \).
2. The foot of the perpendicular from the centre on a tangent lies on the pedal curve \( (x^2+y^2)^2=a^2x^2+b^2y^2 \), which is \( (x^2+y^2)^2=6x^2+2y^2 \).

Teacher's Note:
1. The pedal curve of the ellipse from its centre is (x²+y²)² = a²x² + b²y².

Answer: (d) \( x^2+4y^2=16 \)

Explanation:
1. The diameters have lengths \( 2 \) and \( 4 \), so the semi-minor axis is \( 2 \) and the semi-major axis is \( 4 \).
2. With the major axis along the x-axis: \( \frac{x^2}{16}+\frac{y^2}{4}=1 \), that is \( x^2+4y^2=16 \).

Teacher's Note:
1. Read the semi-axis lengths from the circle diameters and write the standard form.

Answer: (d) \( 2ab \) sq. unit

Explanation:
1. A vertex at \( (a\cos\theta,b\sin\theta) \) gives rectangle area \( 4ab\sin\theta\cos\theta=2ab\sin2\theta \).
2. The maximum is \( 2ab \), at \( \theta=45^\circ \).

Teacher's Note:
1. Express the area through the parameter θ and maximise sin 2θ.

Answer: (d) \( \frac1{\sqrt2} \)

Explanation:
1. Since \( \angle FBF'=90^\circ \) and \( OB \) bisects it, \( \angle OBF=45^\circ \), so \( OB=OF \), i.e. \( b=ae \).
2. \( b^2=a^2(1-e^2)=a^2e^2 \) gives \( e^2=\frac12 \) and \( e=\frac1{\sqrt2} \).

Teacher's Note:
1. OB is the median and the bisector, so b = c.

Answer: (c) \( 13 \)

Explanation:
1. \( F_1F_2 \) is a diameter of the circle, so \( \angle F_1PF_2=90^\circ \). Then \( PF_1^2+PF_2^2=4c^2 \) and \( PF_1+PF_2=17 \).
2. Area \( =\frac12PF_1\cdot PF_2=30 \), so \( PF_1\cdot PF_2=60 \). Then \( 4c^2=17^2-2(60)=169 \).
3. \( 2c=13 \).

Teacher's Note:
1. A circle on the focal segment as diameter makes the angle at P a right angle.

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

Explanation:
1. The ellipse touches the circle at the ends of its major axis, so the circle has radius \( a \). Probability inside \( =\frac{\pi ab}{\pi a^2}=\frac13 \), so \( \frac ba=\frac13 \).
2. \( e^2=1-\frac{b^2}{a^2}=\frac89 \), so \( e=\frac{2\sqrt2}{3} \).

Teacher's Note:
1. Probability here is a ratio of areas; the circle has radius equal to the semi-major axis.

Answer: (a) \( \frac{(a^2-b^2)ab}{a^2+b^2} \)

Explanation:
1. Tangent: \( \frac{x}{a\sqrt2}+\frac{y}{b\sqrt2}=1 \), distance from the centre \( d_1=\frac{\sqrt2\,ab}{\sqrt{a^2+b^2}} \).
2. Normal: \( \sqrt2ax-\sqrt2by=a^2-b^2 \), distance \( d_2=\frac{a^2-b^2}{\sqrt2\sqrt{a^2+b^2}} \).
3. Area \( =d_1d_2=\frac{(a^2-b^2)ab}{a^2+b^2} \).

Teacher's Note:
1. The two perpendiculars from the centre are perpendicular to each other, so their lengths are the sides of the rectangle.

Answer: (a) \( (1+\sin^2\theta)^{-1/2} \)

Explanation:
1. A chord of the circle \( x^2+y^2=a^2 \) subtending \( 90^\circ \) at the centre is at distance \( \frac{a}{\sqrt2} \) from it.
2. Distance of the tangent from the centre: \( \frac{1}{\sqrt{\cos^2\theta/a^2+\sin^2\theta/b^2}}=\frac a{\sqrt2} \), so \( \cos^2\theta+\frac{a^2}{b^2}\sin^2\theta=2 \) and \( \frac{b^2}{a^2}=\frac{\sin^2\theta}{1+\sin^2\theta} \).
3. \( e^2=1-\frac{b^2}{a^2}=\frac1{1+\sin^2\theta} \).

Teacher's Note:
1. Use the distance of the tangent from the centre and the right-angle condition on the auxiliary circle.

Answer: (b) \( abe \)

Explanation:
1. The base \( F_1F_2=2ae \) is fixed. The height is the \( y \)-coordinate of \( P \), at most \( b \).
2. Maximum area \( =\frac12(2ae)(b)=abe \).

Teacher's Note:
1. With a fixed base, the area is largest when P is at an end of the minor axis.

Answer: (d) \( 4 \)

Explanation:
1. \( \tan\frac B2\tan\frac C2=\frac{s-a}{s} \) with \( a=BC=6 \). So \( \frac{s-6}{s}=\frac14 \) and \( s=8 \), i.e. \( AB+AC=10 \).
2. \( A \) lies on an ellipse with foci \( B \), \( C \) and \( 2a=10 \): centre \( (5,0) \), \( a=5 \), \( c=3 \), \( b^2=16 \).
3. Hence \( k=4 \).

Teacher's Note:
1. The half-angle tangent product gives a constant perimeter, which defines an ellipse with foci B and C.

Answer: (a) \( \frac{be}{1+e} \)

Explanation:
1. The area is maximum when \( P \) is an end of the minor axis: area \( =\frac12(2ae)b=abe \).
2. The semi-perimeter is \( s=\frac{2ae+2a}{2}=a(1+e) \).
3. Inradius \( r=\frac{\text{area}}{s}=\frac{be}{1+e} \).

Teacher's Note:
1. Inradius equals area divided by semi-perimeter.

Answer: (c) \( \frac{68}{15} \)

Explanation:
1. Tangent: \( 6x+10y=32 \), so \( Q=\left(\frac{16}{3},0\right) \). Normal: \( \frac{x-2}{6}=\frac{y-2}{10} \), so \( R=\left(\frac45,0\right) \).
2. \( QR=\frac{16}{3}-\frac45=\frac{68}{15} \) and the height from \( P \) is 2.
3. Area \( =\frac12\cdot\frac{68}{15}\cdot2=\frac{68}{15} \).

Teacher's Note:
1. Find the x-intercepts of the tangent and normal, then use base times height.

Answer: (c) \( \frac1{\sqrt2} \)

Explanation:
1. For \( (a\cos t,b\sin t) \): \( r^2=b^2+(a^2-b^2)\cos^2t \). Setting \( r^2=\frac{a^2+2b^2}{2} \) gives \( \cos^2t=\frac{a^2}{2(a^2-b^2)} \).
2. Exactly two points occur only when \( \cos^2t=1 \) (the end points of the major axis): \( a^2=2(a^2-b^2) \), i.e. \( a^2=2b^2 \).
3. \( e^2=1-\frac{b^2}{a^2}=\frac12 \), so \( e=\frac1{\sqrt2} \).

Teacher's Note:
1. Count the solutions of the distance equation in the parameter t; two solutions arise only at the vertices.

Answer: (c) \( (2,1) \)

Explanation:
1. The ellipse is \( \frac{x^2}{6}+\frac{y^2}{3}=1 \). The nearest point has a tangent parallel to \( x+y=7 \), i.e. slope \( -1 \).
2. Tangents of slope \( -1 \): \( y=-x\pm3 \). The nearer one is \( y=-x+3 \), touching at \( \left(\frac{-a^2m}{c},\frac{b^2}{c}\right)=(2,1) \).

Teacher's Note:
1. The nearest point on the ellipse is where the tangent is parallel to the given line.

Answer: (a) \( 4 \)

Explanation:
1. Take \( P(3\cos\theta,2\sin\theta) \), \( A=(-3,0) \). The tangent meets the y-axis at \( Q=\left(0,\frac{2}{\sin\theta}\right) \).
2. Chord \( AP \) meets the y-axis at \( M=\left(0,\frac{2\sin\theta}{1+\cos\theta}\right) \), so \( OQ^2-MQ^2=OM(2OQ-OM) \).
3. \( 2OQ-OM=\frac{2(1+\cos\theta)}{\sin\theta} \), hence the product is \( \frac{2\sin\theta}{1+\cos\theta}\cdot\frac{2(1+\cos\theta)}{\sin\theta}=4 \) for every \( \theta \).

Teacher's Note:
1. The difference of squares simplifies to the constant b².

Answer: (d) \( \ge8 \)

Explanation:
1. The product \( S_1F_1\cdot S_2F_2=b^2=16 \).
2. By AM-GM, \( S_1F_1+S_2F_2\ge2\sqrt{16}=8 \).

Teacher's Note:
1. The product of focal perpendiculars is b², so AM-GM gives the least sum.

Answer: (c) 0

Explanation:
1. The point from which perpendicular tangents are drawn lies on the director circle \( x^2+y^2=16+25=41 \).
2. For \( (6,7) \): \( 36+49=85 \gt 41 \), so the point is outside the director circle and the tangents from it make an acute angle. No perpendicular pair exists: 0.

Teacher's Note:
1. Perpendicular tangents can be drawn only from points on the director circle.