JEE Mathematics Ellipse MCQs Set 05

Answer: (d) \( \dfrac2{17} \)

Explanation:
1. Tangent at \( (1,2) \): \( 2x+y=4 \) (slope \( -2 \)). Tangent at \( (a,b) \): \( 4ax+by=8 \) (slope \( -4a/b \)). Perpendicular: \( \left(-2\right)\left(-\dfrac{4a}b\right)=-1 \), so \( b=-8a \).
2. \( (a,b) \) on the ellipse: \( 4a^2+64a^2=8 \), so \( a^2=\dfrac8{68}=\dfrac2{17} \). Answer: (d)

Teacher's Note:
1. Use the tangent form \( T=0 \) and the perpendicularity condition.

Answer: (a) \( 5 \)

Explanation:
1. Let the semi-axes be \( b\gt a \) (major axis along \( y \)): \( b-a=5 \) and \( be=5\sqrt3 \), so \( b^2-a^2=75 \), giving \( b+a=15 \), \( b=10 \), \( a=5 \).
2. Latus rectum \( =\dfrac{2a^2}b=5 \). Answer: (a)

Teacher's Note:
1. With the major axis on the y-axis, \( a^2=b^2(1-e^2) \).

Answer: (b) \( 9 \)

Explanation:
1. The tangent at \( \left(3,-\frac92\right) \) is \( \dfrac{3x}{a^2}-\dfrac{9y}{2b^2}=1 \). Comparing with \( \dfrac x{12}-\dfrac y6=1 \): \( a^2=36 \) and \( b^2=27 \).
2. Latus rectum \( =\dfrac{2b^2}a=\dfrac{54}6=9 \). Answer: (b)

Teacher's Note:
1. Match the coefficients of the tangent at a given point.

Answer: (d) \( \dfrac{68}{15} \)

Explanation:
1. Tangent: \( 6x+10y=32 \), meeting the axis at \( Q\left(\dfrac{16}3,0\right) \). Normal: slope \( \dfrac53 \), \( y-2=\dfrac53(x-2) \), meeting the axis at \( R\left(\dfrac45,0\right) \).
2. \( QR=\dfrac{16}3-\dfrac45=\dfrac{68}{15} \) and the height is \( 2 \), so the area is \( \dfrac12\cdot\dfrac{68}{15}\cdot2=\dfrac{68}{15} \). Answer: (d)

Teacher's Note:
1. The slope of the normal is the negative reciprocal of the tangent slope.

Answer: (a) \( \dfrac{5\sqrt5}2 \)

Explanation:
1. Let \( P(2\cos\theta,\sqrt3\sin\theta) \). The normal has slope \( -2 \), so the tangent slope is \( \dfrac12 \), which gives \( \tan\theta=-\sqrt3 \), \( P\left(-1,\dfrac32\right) \) (the tangent \( -\frac x4+\frac y2=1 \) passes through \( Q(4,4) \)).
2. \( PQ=\sqrt{5^2+\left(\dfrac52\right)^2}=\dfrac{5\sqrt5}2 \). Answer: (a)

Teacher's Note:
1. Find \( P \) from the slope of the normal, then use the distance formula.

Answer: (a) \( (\sqrt2,2) \)

Explanation:
1. The major axis is along the \( y \)-axis with \( be=2 \) (semi-minor axis \( 2 \)), so \( b^2=a^2+4=8 \) in the form \( \dfrac{x^2}4+\dfrac{y^2}8=1 \).
2. Only \( (\sqrt2,2) \) satisfies it: \( \dfrac24+\dfrac48=1 \). Answer: (a)

Teacher's Note:
1. \( c^2=b^2-a^2 \) with the major semi-axis along \( y \).

Answer: (d) \( \dfrac1{2x^2}+\dfrac1{4y^2}=1 \)

Explanation:
1. The ellipse is \( \dfrac{x^2}2+y^2=1 \); the tangent at \( (\sqrt2\cos\theta,\sin\theta) \) is \( \dfrac{x\cos\theta}{\sqrt2}+y\sin\theta=1 \), meeting the axes at \( (\sqrt2\sec\theta,0) \) and \( (0,\mathrm{cosec}\,\theta) \).
2. The mid-point \( (h,k) \) has \( \cos\theta=\dfrac1{\sqrt2h} \), \( \sin\theta=\dfrac1{2k} \), so \( \dfrac1{2h^2}+\dfrac1{4k^2}=1 \). Answer: (d)

Teacher's Note:
1. Eliminate the parameter using \( \sin^2\theta+\cos^2\theta=1 \).

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

Explanation:
1. \( 2ae=2b \) gives \( e=\dfrac ba \) and \( e^2=1-e^2 \), so \( e=\dfrac1{\sqrt2} \). Latus rectum \( \dfrac{2b^2}a=8 \) gives \( b=4\sqrt2 \) and \( a=8 \).
2. The ellipse is \( \dfrac{x^2}{64}+\dfrac{y^2}{32}=1 \), which is satisfied by \( (4\sqrt3,2\sqrt2) \): \( \dfrac{48}{64}+\dfrac8{32}=1 \). Answer: (b)

Teacher's Note:
1. Distance between foci \( 2ae \) equals \( 2b \).

Answer: (d) \( 4 \)

Explanation:
1. \( SB\perp S'B \) with \( SB=S'B=a \) gives \( b=ae \) and \( e^2=\dfrac12 \). Area \( =\dfrac12a^2=8 \), so \( a=4 \).
2. \( b^2=a^2e^2=8 \), and the latus rectum is \( \dfrac{2b^2}a=4 \). Answer: (d)

Teacher's Note:
1. The distance from a minor-axis end to a focus is \( a \).

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

Explanation:
1. \( \dfrac ae=4 \) and \( e=\dfrac12 \) give \( a=2 \), \( b^2=a^2(1-e^2)=3 \): \( \dfrac{x^2}4+\dfrac{y^2}3=1 \).
2. The normal at \( (x_1,y_1) \) is \( \dfrac{a^2x}{x_1}-\dfrac{b^2y}{y_1}=a^2-b^2 \): \( 4x-2y=1 \). Answer: (b)

Teacher's Note:
1. Use the normal formula for an ellipse.

Answer: (d) \( 27 \)

Explanation:
1. \( e=\dfrac23 \), foci \( (\pm2,0) \), and the ends of a latus rectum are \( \left(2,\pm\dfrac53\right) \). The tangent at \( \left(2,\dfrac53\right) \) is \( \dfrac{2x}9+\dfrac y3=1 \), i.e. \( 2x+3y=9 \), meeting the axes at \( \left(\dfrac92,0\right) \) and \( (0,3) \).
2. By symmetry the quadrilateral is made of four congruent triangles: area \( =4\times\dfrac12\cdot\dfrac92\cdot3=27 \). Answer: (d)

Teacher's Note:
1. The quadrilateral is a rhombus symmetric about both axes.

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

Explanation:
1. For the ellipse \( \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 \) the foot \( (h,k) \) of the perpendicular from the centre on a tangent satisfies \( (h^2+k^2)^2=a^2h^2+b^2k^2 \).
2. Here \( a^2=6 \), \( b^2=2 \), so the locus is \( (x^2+y^2)^2=6x^2+2y^2 \). Answer: (c)

Teacher's Note:
1. This is the pedal curve of the ellipse with respect to its centre.

Answer: (d) \( 3x^2+5y^2-32=0 \)

Explanation:
1. \( e^2=1-\dfrac{b^2}{a^2}=\dfrac25 \) gives \( \dfrac{b^2}{a^2}=\dfrac35 \). The point: \( \dfrac9{a^2}+\dfrac1{b^2}=1 \), i.e. \( \dfrac9{a^2}+\dfrac5{3a^2}=1 \), so \( a^2=\dfrac{32}3 \), \( b^2=\dfrac{32}5 \).
2. The ellipse is \( \dfrac{3x^2}{32}+\dfrac{5y^2}{32}=1 \), i.e. \( 3x^2+5y^2-32=0 \). Answer: (d)

Teacher's Note:
1. Use \( e^2=1-b^2/a^2 \) and the given point.

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

Explanation:
1. The rectangle has sides \( x=\pm2 \) and \( y=\pm1 \), so its corner \( (2,1) \) lies on the outer ellipse \( \dfrac{x^2}{16}+\dfrac{y^2}{b^2}=1 \) (since it passes through \( (4,0) \)).
2. \( \dfrac14+\dfrac1{b^2}=1 \) gives \( b^2=\dfrac43 \), so the ellipse is \( \dfrac{x^2}{16}+\dfrac{3y^2}4=1 \), i.e. \( x^2+12y^2=16 \). Answer: (a)

Teacher's Note:
1. The rectangle circumscribing the inner ellipse has its corners on the outer ellipse.

Answer: (b) \( \dfrac83 \)

Explanation:
1. The distance between the directrix and the nearer focus is \( \dfrac ae-ae=4 \).
2. With \( e=\dfrac12 \): \( 2a-\dfrac a2=4 \), so \( a=\dfrac83 \). Answer: (b)

Teacher's Note:
1. Focus-to-directrix distance is \( a/e-ae \).

Answer: (d) \( \dfrac35 \)

Explanation:
1. \( 2ae=6 \) gives \( ae=3 \) and \( 2b=8 \) gives \( b=4 \), so \( a^2=b^2+(ae)^2=25 \), \( a=5 \).
2. \( e=\dfrac{ae}a=\dfrac35 \). Answer: (d)

Teacher's Note:
1. Use \( a^2=b^2+c^2 \) with \( c=ae \).

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

Explanation:
1. A rectangle with vertices \( (\pm a\cos\theta,\pm b\sin\theta) \) has area \( 4ab\sin\theta\cos\theta=2ab\sin2\theta \).
2. The maximum is at \( \theta=\dfrac\pi4 \), where the area is \( 2ab \). Answer: (d)

Teacher's Note:
1. Parametrise the vertices of the rectangle.

Answer: (d) \( \dfrac1{\sqrt2} \)

Explanation:
1. Since \( \angle FBF'=90^\circ \) and \( BF=BF' \), \( \angle OBF=45^\circ \), so \( OF=OB \), i.e. \( ae=b \).
2. Then \( e^2=1-\dfrac{b^2}{a^2}=1-e^2 \), so \( e=\dfrac1{\sqrt2} \). Answer: (d)

Teacher's Note:
1. The triangle \( OBF \) is an isosceles right triangle.

Answer: (b) \( 3x^2+4y^2=12 \)

Explanation:
1. The directrix \( x=\dfrac ae=4 \) with \( e=\dfrac12 \) gives \( a=2 \), and \( b^2=a^2(1-e^2)=3 \).
2. The ellipse is \( \dfrac{x^2}4+\dfrac{y^2}3=1 \), i.e. \( 3x^2+4y^2=12 \). Answer: (b)

Teacher's Note:
1. A directrix of the form \( x=c \) means the major axis lies along the x-axis.

Answer: (b) \( \dfrac{x^2}{16}+\dfrac{y^2}{12}=1 \)

Explanation:
1. \( ae=2 \) and \( e=\dfrac12 \) give \( a=4 \), so \( b^2=a^2(1-e^2)=16\cdot\dfrac34=12 \).
2. The ellipse is \( \dfrac{x^2}{16}+\dfrac{y^2}{12}=1 \). Answer: (b)

Teacher's Note:
1. Foci on the x-axis give the major axis along the x-axis.