JEE Mathematics Ellipse and Hyperbola MCQs Set 04

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

Explanation:
1. The directrix is \( x=\frac ae=4 \), so \( a=2 \).
2. \( b^2=a^2(1-e^2)=3 \).
3. Equation \( \frac{x^2}4+\frac{y^2}3=1 \), i.e. \( 3x^2+4y^2=12 \).

Teacher's Note:
1. The directrices are x = +- a/e.
2. Then use b squared = a squared (1 - e squared).

Answer: (b) \( 9x^2+y^2-6xy-54x-62y+241=0 \)

Explanation:
1. Line \( AB:\ x+3y-3=0 \). Equating squared distances: \( (x-3)^2+(y-4)^2=\frac{(x+3y-3)^2}{10} \).
2. Expanding: \( 9x^2+y^2-6xy-54x-62y+241=0 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( x^2-3y^2-2x+1=0 \)

Explanation:
1. In standard form \( (x-1)^2-3y^2=9 \).
2. The asymptotes are obtained by replacing 9 by 0: \( (x-1)^2-3y^2=0 \).
3. Expanding: \( x^2-3y^2-2x+1=0 \).

Teacher's Note:
1. Asymptotes differ from the hyperbola only in the constant term.
2. They pass through the centre (1, 0).

Answer: (d) \( (3,0)\text{ and }\left(-\frac95,\frac85\right) \)

Explanation:
1. Chord of contact: \( \frac{3x}9+\frac{4y}4=1 \), i.e. \( y=1-\frac x3 \).
2. Substituting: \( 5x^2-6x-27=0 \), so \( x=3 \) or \( x=-\frac95 \), giving \( (3,0) \) and \( (-\frac95,\frac85) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( x^2-3y^2-2x+10=0 \)

Explanation:
1. Complete the square: \( (x-1)^2-3y^2=9 \), i.e. \( \frac{(x-1)^2}{9}-\frac{y^2}{3}=1 \).
2. The conjugate is \( \frac{(x-1)^2}{9}-\frac{y^2}{3}=-1 \), i.e. \( (x-1)^2-3y^2=-9 \).
3. Expanding: \( x^2-3y^2-2x+10=0 \).

Teacher's Note:
1. Conjugate hyperbola: replace 1 by -1 on the right in standard form.
2. Always complete the square first for a shifted hyperbola.

Answer: (a) \( \frac45 \)

Explanation:
1. The foci are \( \pm4 \) and \( 2a=10 \), so \( a=5,\ c=4 \).
2. Eccentricity \( e=\frac ca=\frac45 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. The hyperbola \( \frac{x^2}{1/2}-\frac{y^2}{1/2}=1 \) has \( e=\sqrt2 \) and foci \( (\pm1,0) \).
2. Orthogonal intersection implies confocal; the ellipse has \( e=\frac1{\sqrt2} \) and \( ae=1 \), so \( a^2=2,\ b^2=1 \).
3. Ellipse: \( x^2+2y^2=2 \) with foci \( (\pm1,0) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \left(\frac{11}5,\frac85\right) \)

Explanation:
1. Here \( A(3,0),\ B(-\frac95,\frac85) \). \( PA \) is vertical, so the altitude from \( B \) is \( y=\frac85 \).
2. Slope of \( AB \) is \( -\frac13 \), so the altitude from \( P \) is \( y-4=3(x-3) \). With \( y=\frac85 \): \( x=\frac{11}5 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \left(\pm2\sqrt3,\pm\frac17\right) \)

Explanation:
1. With \( P(4\cos\phi,2\sin\phi) \) the normal is \( \frac{4x}{\cos\phi}-\frac{2y}{\sin\phi}=12 \), meeting the x-axis at \( Q(3\cos\phi,0) \).
2. Midpoint \( M\left(\tfrac72\cos\phi,\sin\phi\right) \): locus \( \frac{4x^2}{49}+y^2=1 \).
3. Latus rectum \( x=\pm2\sqrt3 \): \( y^2=1-\frac{48}{49}=\frac1{49} \), so \( y=\pm\frac17 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. The hyperbola is \( \frac{x^2}{18}-\frac{y^2}{9}=1 \). A tangent \( y=mx+c \) needs \( c^2=a^2m^2-b^2 \).
2. With \( m=1 \): \( c^2=18-9=9 \), so \( c=\pm3 \).
3. Tangents: \( y=x+3 \) and \( y=x-3 \), i.e. (i) and (iv).

Teacher's Note:
1. Use the condition c squared = a squared m squared - b squared.
2. Parallel to x - y = 0 means slope 1.

Answer: (c) \( \frac{27}{10} \)

Explanation:
1. \( A(3,0),\ B(0,1) \): line \( \frac x3+y=1 \); auxiliary circle \( x^2+y^2=9 \).
2. Substituting \( x=3(1-y) \): \( 10y^2-18y=0 \), so \( y=\frac95 \), \( x=-\frac{12}5 \).
3. Area \( =\frac12\cdot3\cdot\frac95=\frac{27}{10} \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. The ellipse is \( \frac{x^2}{4}+\frac{y^2}3=1 \) with \( a=2 \).
2. The foot of the perpendicular from a focus to any tangent lies on the auxiliary circle \( x^2+y^2=a^2 \).
3. So the locus is \( x^2+y^2=4 \).

Teacher's Note:
1. A standard property: feet of focal perpendiculars to tangents lie on the auxiliary circle.
2. The same holds for either focus.

Answer: (c) \( x^2\operatorname{cosec}^2\theta-y^2\sec^2\theta=1 \)

Explanation:
1. The ellipse \( \frac{x^2}4+\frac{y^2}3=1 \) has foci \( (\pm1,0) \).
2. The hyperbola has \( a=\sin\theta,\ c=1 \), so \( b^2=1-\sin^2\theta=\cos^2\theta \).
3. Equation \( \frac{x^2}{\sin^2\theta}-\frac{y^2}{\cos^2\theta}=1 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. The ellipse has \( e=\frac35 \) and foci \( (\pm3,0) \). The hyperbola has \( e=\frac53 \) and passes through \( (3,0) \), so \( a=3 \).
2. \( b^2=a^2(e^2-1)=9\cdot\frac{16}9=16 \): \( \frac{x^2}9-\frac{y^2}{16}=1 \).
3. Foci: \( (\pm ae,0)=(\pm5,0) \). So (i) and (iii) are correct.

Teacher's Note:
1. The eccentricity of the hyperbola is the reciprocal of the ellipse's.
2. The vertex of the hyperbola is at the focus of the ellipse.

Answer: (b) 2

Explanation:
1. The nearest directrix meets the x-axis at \( (\frac ae,0) \); the line passes through it: \( \frac{2a}e=1 \), so \( a=\frac e2 \).
2. Tangency of \( y=-2x+1 \): \( 1=4a^2-b^2=a^2(5-e^2) \), so \( \frac{e^2}4(5-e^2)=1 \) i.e. \( e^4-5e^2+4=0 \).
3. Hence \( e^2=4 \) (since \( e>1 \) and \( e^2=1 \) is rejected): \( e=2 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \sqrt{\frac32}-1 \)

Explanation:
1. Complete the squares: \( (x-\sqrt2)^2-2(y+\sqrt2)^2=4 \), i.e. \( \frac{(x-\sqrt2)^2}{4}-\frac{(y+\sqrt2)^2}{2}=1 \): \( a=2,\ b^2=2,\ e=\sqrt{\frac32} \).
2. Take the right-hand vertex and focus: \( AS=a(e-1)=2\left(\sqrt{\tfrac32}-1\right) \) and the semi latus rectum \( SB=\frac{b^2}{a}=1 \).
3. Area \( =\frac12\cdot AS\cdot SB=\sqrt{\tfrac32}-1 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.