JEE Mathematics Ellipse MCQs Set 08

Answer: (c) \( 4 \)

Explanation:
1. Right angle at \( B \) means \( b=c \), so \( e=\frac1{\sqrt2} \) and \( a^2=2b^2 \).
2. Area \( =\frac12(2c)(b)=b^2=8 \), so \( a^2=16 \), \( a=4 \).
3. Latus rectum \( =\frac{2b^2}{a}=\frac{16}{4}=4 \).

Teacher's Note:
1. A right angle at the end of the minor axis forces b = c, that is e = 1/√2.

Answer: (a) \( x^2+12y^2=16 \)

Explanation:
1. The inner ellipse \( \frac{x^2}{4}+y^2=1 \) touches the rectangle \( x=\pm2 \), \( y=\pm1 \), whose corners are \( (\pm2,\pm1) \).
2. The outer ellipse \( \frac{x^2}{16}+\frac{y^2}{B}=1 \) passes through \( (2,1) \): \( \frac{4}{16}+\frac1B=1 \), so \( B=\frac43 \).
3. Hence \( \frac{x^2}{16}+\frac{3y^2}{4}=1 \), i.e. \( x^2+12y^2=16 \).

Teacher's Note:
1. The rectangle's corners lie on the outer ellipse, which also passes through (4, 0).

Answer: (d) \( \frac35 \)

Explanation:
1. \( 2c=6 \) gives \( c=3 \); \( 2b=8 \) gives \( b=4 \).
2. \( a^2=b^2+c^2=25 \), so \( a=5 \) and \( e=\frac ca=\frac35 \).

Teacher's Note:
1. Use a² = b² + c² to find the semi-major axis first.

Answer: (c) a point if \( k=0 \)

Explanation:
1. Complete the squares: \( 2(x-2)^2+3(y-3)^2=k \).
2. For \( k \gt 0 \) it is an ellipse, for \( k=0 \) only the point \( (2,3) \), and for \( k \lt 0 \) there is no locus. So only option (c) is correct.

Teacher's Note:
1. Complete the squares to see how the right-hand side decides the nature of the curve.

Answer: (c) \( (0,4) \) and \( (0,1) \)

Explanation:
1. For a mid-point \( (h,k) \), \( T=S_1 \) gives \( \frac{hx}{16}+\frac{ky}{25}-1=\frac{h^2}{16}+\frac{k^2}{25}-1 \). The chord passes through \( (0,5) \): \( \frac{k}{5}=\frac{h^2}{16}+\frac{k^2}{25} \).
2. So \( \frac{x^2}{16}+\frac{y^2}{25}=\frac y5 \), i.e. \( \frac{x^2}{16}+\frac{(y-5/2)^2}{25}=\frac14 \), which is \( \frac{x^2}{4}+\frac{(y-5/2)^2}{25/4}=1 \) with centre \( \left(0,\frac52\right) \).
3. \( c^2=\frac{25}{4}-4=\frac94 \), so \( c=\frac32 \) and the foci are \( \left(0,\frac52\pm\frac32\right)=(0,4) \) and \( (0,1) \).

Teacher's Note:
1. Use the mid-point chord equation T = S₁ and complete the square.

Answer: (c) \( b^2 \)

Explanation:
1. For the tangent \( y=mx+\sqrt{a^2m^2+b^2} \), the distances from \( (\pm ae,0) \) are \( \frac{|\pm aem+c|}{\sqrt{1+m^2}} \).
2. Their product is \( \frac{c^2-a^2e^2m^2}{1+m^2}=\frac{b^2(1+m^2)}{1+m^2}=b^2 \).

Teacher's Note:
1. The product of focal perpendiculars on a tangent is the constant b².

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

Explanation:
1. The tangents at \( \alpha \) and \( \beta \) meet at \( \left(\frac{a\cos\frac{\alpha+\beta}2}{\cos\frac{\alpha-\beta}2},\frac{b\sin\frac{\alpha+\beta}2}{\cos\frac{\alpha-\beta}2}\right) \).
2. So \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{\cos^2\frac{\alpha-\beta}2}=\frac1{\cos^2 45^\circ}=2 \).

Teacher's Note:
1. Use the coordinates of the pole of the chord joining two eccentric angles.

Answer: (c) an ellipse

Explanation:
1. Take \( P(3\cos\theta,\sin\theta) \) and \( Q(3\cos\theta,-\sin\theta) \). The normal at \( P \) is \( \frac{3x}{\cos\theta}-\frac{y}{\sin\theta}=8 \) and the diameter \( OQ \) is \( y=-\frac{\tan\theta}{3}x \).
2. Solving gives \( R=\left(\frac{12\cos\theta}{5},-\frac{4\sin\theta}{5}\right) \) and the mid-point of \( PR \) is \( \left(\frac{27\cos\theta}{10},\frac{\sin\theta}{10}\right) \).
3. Eliminating \( \theta \): \( \left(\frac{10x}{27}\right)^2+(10y)^2=1 \), an ellipse.

Teacher's Note:
1. Parametrise P, find R and eliminate the parameter from the mid-point.

Answer: (b) \( e^4+e^2-1=0 \)

Explanation:
1. The end of the latus rectum is \( \left(ae,\frac{b^2}{a}\right) \); its normal is \( \frac{a^2x}{ae}-\frac{b^2y}{b^2/a}=a^2e^2 \).
2. This simplifies to \( \frac{ax}{e}-ay=a^2e^2 \). Putting \( (0,-b) \): \( ab=a^2e^2 \), so \( b=ae^2 \).
3. \( b^2=a^2(1-e^2)=a^2e^4 \) gives \( e^4+e^2-1=0 \).

Teacher's Note:
1. Substitute the extremity of the minor axis into the equation of the normal.

Answer: (c) \( \frac1{\sqrt5} \)

Explanation:
1. The tangent is \( x-ty+t^2=0 \). The normal is \( \frac{\sqrt5x}{\cos\theta}-\frac{2y}{\sin\theta}=1 \).
2. Matching coefficients: \( \cos\theta=-\sqrt5t^2 \) and \( \sin\theta=-2t \). Then \( 5t^4+4t^2=1 \), i.e. \( (5t^2-1)(t^2+1)=0 \) and \( t^2=\frac15 \).
3. The value in the options is \( t=\frac1{\sqrt5} \) (then \( \cos\theta=-\frac1{\sqrt5} \), \( \sin\theta=-\frac2{\sqrt5} \) is valid).

Teacher's Note:
1. Equate the ratios of the coefficients of the two line equations.

Answer: (b) \( \frac{a^2}{x^2}+\frac{b^2}{y^2}=4 \)

Explanation:
1. The tangent at \( (a\cos\theta,b\sin\theta) \) meets the axes at \( A\left(\frac a{\cos\theta},0\right) \) and \( B\left(0,\frac b{\sin\theta}\right) \).
2. Mid-point: \( x=\frac{a}{2\cos\theta} \), \( y=\frac{b}{2\sin\theta} \), so \( \cos^2\theta=\frac{a^2}{4x^2} \) and \( \sin^2\theta=\frac{b^2}{4y^2} \).
3. Adding: \( \frac{a^2}{x^2}+\frac{b^2}{y^2}=4 \).

Teacher's Note:
1. Find the intercepts of the tangent in parametric form and eliminate θ.

Answer: (b) \( \frac2{17} \)

Explanation:
1. The tangent at \( (1,2) \) is \( 4x+2y=8 \), with slope \( -2 \). The perpendicular tangent has slope \( \frac12 \).
2. The tangent at \( (a,b) \) is \( 4ax+by=8 \) with slope \( -\frac{4a}{b}=\frac12 \), so \( b=-8a \).
3. On the ellipse: \( 4a^2+64a^2=8 \), so \( a^2=\frac{2}{17} \).

Teacher's Note:
1. Use the slope of the tangent at a point and the condition for perpendicular lines.

Answer: (c) \( 5 \)

Explanation:
1. \( 2a-2b=10 \) gives \( a-b=5 \). The focus on the y-axis gives \( c=5\sqrt3 \), so \( a^2-b^2=75 \).
2. Then \( a+b=15 \), so \( a=10 \), \( b=5 \).
3. Latus rectum \( =\frac{2b^2}{a}=\frac{50}{10}=5 \).

Teacher's Note:
1. Combine a − b and a² − b² = c² to get a + b.

Answer: (c) \( \sqrt2+1 \)

Explanation:
1. The tangent at \( (\alpha,\beta) \) is \( y=\frac{x}{2\beta}+\frac\beta2 \) (using \( \alpha=\beta^2 \)).
2. For the ellipse \( \frac{x^2}{1}+\frac{y^2}{1/2}=1 \) the condition is \( c^2=a^2m^2+b^2 \): \( \frac{\beta^2}{4}=\frac{1}{4\beta^2}+\frac12 \).
3. \( \beta^4-2\beta^2-1=0 \), so \( \beta^2=1+\sqrt2 \) and \( \alpha=\beta^2=\sqrt2+1 \).

Teacher's Note:
1. Apply the tangency condition c² = a²m² + b² to the parabola's tangent line.

Answer: (a) \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac12 \)

Explanation:
1. Mid-point: \( x=\frac a2(\cos\theta-\sin\theta) \), \( y=\frac b2(\sin\theta+\cos\theta) \).
2. \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac14\left[(\cos\theta-\sin\theta)^2+(\sin\theta+\cos\theta)^2\right]=\frac12 \).

Teacher's Note:
1. Add the coordinates of P and Q and halve; the squares then collapse using sin² + cos² = 1.

Answer: (c) \( \pm2 \)

Explanation:
1. Perpendicular tangents meet on the director circle \( x^2+y^2=a^2+b^2=13 \).
2. \( \lambda^2+9=13 \), so \( \lambda=\pm2 \).

Teacher's Note:
1. The director circle of an ellipse is x² + y² = a² + b².

Answer: (c) \( \frac\pi4 \)

Explanation:
1. The normal at \( \theta \) is \( \frac{5x}{\cos\theta}-\frac{3y}{\sin\theta}=16 \).
2. Comparing with \( 5x-3y=8\sqrt2 \): \( \frac1{\cos\theta}=\frac1{\sin\theta}=\frac{16}{8\sqrt2}=\sqrt2 \), so \( \theta=\frac\pi4 \).

Teacher's Note:
1. Compare the normal at the parametric point with the given line.

Answer: (c) \( \frac{\sqrt{43}}{5} \)

Explanation:
1. A tangent \( y-2=m(x-4) \) satisfies \( (2-4m)^2=9m^2+1 \), i.e. \( 7m^2-16m+3=0 \).
2. \( m_1+m_2=\frac{16}{7} \), \( m_1m_2=\frac37 \), so \( |m_1-m_2|=\frac{\sqrt{256-84}}{7}=\frac{2\sqrt{43}}{7} \).
3. \( \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|=\frac{2\sqrt{43}/7}{10/7}=\frac{\sqrt{43}}{5} \).

Teacher's Note:
1. Find the two tangent slopes from the quadratic and use the angle formula.

Answer: (a) \( 12 \)

Explanation:
1. The vertices of a square circumscribing the ellipse lie on the director circle \( x^2+y^2=a^2+b^2 \), so the diagonal is \( 2\sqrt{a^2+b^2} \) and the side is \( \sqrt2\sqrt{a^2+b^2} \).
2. \( \sqrt2\sqrt{16+b^2}=5\sqrt2 \) gives \( b^2=9 \).
3. \( A=\pi ab=12\pi \), so \( \frac A\pi=12 \).

Teacher's Note:
1. A square circumscribing an ellipse has its corners on the director circle.

Answer: (a) \( \frac\pi4 \)

Explanation:
1. Points at distance 2: \( x^2+y^2=4 \). With \( x^2+3y^2=6 \): \( y^2=1 \), \( x^2=3 \).
2. Write \( (\sqrt6\cos\theta,\sqrt2\sin\theta) \): \( \sin\theta=\pm\frac1{\sqrt2} \) and \( \cos\theta=\pm\frac1{\sqrt2} \), giving \( \theta=\frac\pi4 \) among the options.

Teacher's Note:
1. Intersect the ellipse with the circle of radius 2 about the centre.

Answer: (a) \( \cos^{-1}\left(-\frac23\right) \)

Explanation:
1. Normal at \( P \): \( \frac{\sqrt{14}x}{\cos\theta}-\frac{\sqrt5y}{\sin\theta}=9 \). Put in \( Q \): \( \frac{14\cos2\theta}{\cos\theta}-10\cos\theta=9 \).
2. With \( c=\cos\theta \): \( 14(2c^2-1)-10c^2=9c \), i.e. \( 18c^2-9c-14=0 \).
3. \( c=\frac{9\pm33}{36} \); the value \( \frac{42}{36} \gt 1 \) is rejected, so \( \cos\theta=-\frac23 \).

Teacher's Note:
1. Substitute the second point into the equation of the normal and solve the quadratic in cos θ.