JEE Mathematics Ellipse and Hyperbola MCQs Set 06

Answer: (b) \( 7x^2+2xy+7y^2-10x+10y+7=0 \)

Explanation:
1. For a point (x, y): \( PS^2=e^2\,d^2 \), i.e. \( (x-1)^2+(y+1)^2=\frac14\cdot\frac{(x-y-3)^2}{2} \).
2. Multiply by 8: \( 8(x^2-2x+1+y^2+2y+1)=x^2+y^2+9-2xy-6x+6y \).
3. Simplify: \( 7x^2+2xy+7y^2-10x+10y+7=0 \).

Teacher's Note:
1. Use focus-directrix definition: PS = e × distance from the directrix.
2. Expand carefully; the xy coefficient comes from −(x − y − 3)².

Answer: (a) \( \frac{\sqrt5}{3} \)

Explanation:
1. Complete the squares: \( 4(x+1)^2+9(y+2)^2=36 \), i.e. \( \frac{(x+1)^2}{9}+\frac{(y+2)^2}{4}=1 \).
2. So \( a^2=9,\ b^2=4 \) and \( e^2=1-\frac49=\frac59 \).
3. Hence \( e=\frac{\sqrt5}{3} \).

Teacher's Note:
1. Complete the square in x and y separately, then divide by the constant.
2. e² = 1 − b²/a² with a the larger semi-axis.

Answer: (b) \( r\in(2,5)-\{3.5\} \)

Explanation:
1. Rewrite as \( \frac{x^2}{r-2}+\frac{y^2}{5-r}=1 \).
2. Both denominators must be positive: \( r>2 \) and \( r<5 \).
3. They must not be equal (that would be a circle): \( r-2\ne5-r \), so \( r\ne3.5 \).

Teacher's Note:
1. For an ellipse both denominators are positive and unequal.
2. Equal denominators give a circle.

Answer: (d) \( \frac49 \)

Explanation:
1. Write it as \( \frac{x^2}{1/9}+\frac{y^2}{1/4}=1 \). The larger denominator is under y², so the major axis is along y with \( a^2=\frac14 \) and \( b^2=\frac19 \).
2. Latus rectum \( =\frac{2b^2}{a}=\frac{2/9}{1/2}=\frac49 \).

Teacher's Note:
1. Identify which denominator is larger to decide the major axis.
2. Latus rectum = 2b²/a.

Answer: (d) \( \frac{(x-1)^2}{45}+\frac{(y-2)^2}{20}=1 \)

Explanation:
1. The focus is 5 units from the centre along the line y = 2, so the major axis is horizontal and \( a^2-b^2=25 \).
2. The point (4, 6) gives \( \frac{9}{a^2}+\frac{16}{b^2}=1 \).
3. Solving: \( a^2=45,\ b^2=20 \) (check: 9/45 + 16/20 = 1).

Teacher's Note:
1. Use ae = 5 (distance from centre to focus) to get a² − b² = 25.
2. Substitute the given point and solve for a².

Answer: (c) on the major axis

Explanation:
1. Complete the squares: \( 4(x-2)^2+9(y-3)^2=36 \), i.e. \( \frac{(x-2)^2}{9}+\frac{(y-3)^2}{4}=1 \). The centre is (2, 3) and the major axis is horizontal, along y = 3.
2. The point (1, 3) lies on y = 3, i.e. on the major axis (it is also inside the ellipse, as the value of S at the point is 4 + 81 − 16 − 162 + 61 = −32 < 0).
3. It is not on the minor axis x = 2, and not outside or on the curve.

Teacher's Note:
1. The major axis passes through the centre parallel to the longer semi-axis.
2. Only one of the options is true for this point.

Answer: (c) \( x^2+y^2=\frac{7+3\sqrt5}{2} \)

Explanation:
1. The quadratic form has matrix \( \begin{pmatrix}2&3\\3&5\end{pmatrix} \) with eigenvalues \( \lambda=\frac{7\pm3\sqrt5}{2} \).
2. In the principal axes the ellipse is \( \lambda_1X^2+\lambda_2Y^2=1 \), so the semi-axes squared are \( 1/\lambda \).
3. The larger one is \( a^2=\frac{2}{7-3\sqrt5}=\frac{7+3\sqrt5}{2} \). The auxiliary circle is centred at the centre (origin) with radius a: \( x^2+y^2=\frac{7+3\sqrt5}{2} \).

Teacher's Note:
1. Rotate to the principal axes; the xy term disappears and the eigenvalues give the semi-axes.
2. The auxiliary circle has the semi-major axis as radius.

Answer: (a) an ellipse

Explanation:
1. Then \( \frac x3=\cos t+\sin t \) and \( \frac y4=\cos t-\sin t \).
2. Squaring and adding: \( \frac{x^2}{9}+\frac{y^2}{16}=2(\cos^2t+\sin^2t)=2 \).
3. So \( \frac{x^2}{18}+\frac{y^2}{32}=1 \): an ellipse.

Teacher's Note:
1. Square both parametric expressions and add to eliminate t.
2. Unequal denominators mean it is not a circle.

Answer: (b) \( \sqrt2 \)

Explanation:
1. Complete the squares: \( (x-2)^2-(y-2)^2=-16 \).
2. The coefficients of x² and y² are equal in magnitude and opposite in sign: a rectangular hyperbola, a = b.
3. Then \( e=\sqrt{1+b^2/a^2}=\sqrt2 \).

Teacher's Note:
1. A hyperbola with a = b is rectangular.
2. e = √2 for every rectangular hyperbola.

Answer: (c) the length of its transverse axis is \( 4\sqrt3 \)

Explanation:
1. Complete the squares: \( 16(x-1)^2-3(y-2)^2=48 \), i.e. \( \frac{(x-1)^2}{3}-\frac{(y-2)^2}{16}=1 \).
2. So the centre is (1, 2), \( a^2=3,\ b^2=16 \): transverse axis \( 2a=2\sqrt3 \), conjugate axis \( 2b=8 \), \( e^2=1+\frac{16}{3}=\frac{19}3 \).
3. The transverse axis is \( 2\sqrt3 \), not \( 4\sqrt3 \), so the statement giving 4√3 is incorrect.

Teacher's Note:
1. Bring the equation to standard form first.
2. Compare each statement with a² = 3, b² = 16.

Answer: (d) \( \frac2{\sqrt3} \)

Explanation:
1. Conjugate axis = 2b; distance between the foci = 2ae. The condition is \( 2b=ae \).
2. Square: \( 4a^2(e^2-1)=a^2e^2 \), so \( 3e^2=4 \).
3. Hence \( e=\frac2{\sqrt3} \).

Teacher's Note:
1. Use b² = a²(e² − 1).
2. The foci are at distance ae from the centre.

Answer: (d) (4, 1)

Explanation:
1. Complete the squares: \( 9(x+1)^2-16(y-1)^2=144 \), i.e. \( \frac{(x+1)^2}{16}-\frac{(y-1)^2}{9}=1 \).
2. Centre (−1, 1), \( a=4,\ b=3 \), so \( c=\sqrt{16+9}=5 \).
3. The foci are \( (-1\pm5,\ 1) \): (4, 1) and (−6, 1). Only (4, 1) is listed.

Teacher's Note:
1. The transverse axis is horizontal, so the foci have the same y-coordinate as the centre.
2. c² = a² + b² for a hyperbola.

Answer: (c) \( x+y\pm2\sqrt2\,c=0 \)

Explanation:
1. The transverse axis of \( xy=c^2 \) is the line y = x, with foci \( (\sqrt2c,\sqrt2c) \) and \( (-\sqrt2c,-\sqrt2c) \).
2. A latus rectum passes through a focus perpendicular to the transverse axis, so it has slope −1: through the first focus, \( x+y=2\sqrt2c \).
3. The other is \( x+y=-2\sqrt2c \).

Teacher's Note:
1. The foci lie on the line y = x at (±√2c, ±√2c), at distance 2c from the centre.
2. Latus rectum is perpendicular to the transverse axis.

Answer: (d) \( \pm6 \)

Explanation:
1. The tangent condition is \( c^2=a^2m^2+b^2 \).
2. Here \( a^2=8,\ b^2=4,\ m=2 \): \( c^2=32+4=36 \).
3. So \( c=\pm6 \).

Teacher's Note:
1. Tangent condition for x²/a² + y²/b² = 1: c² = a²m² + b².
2. Do not mix up a² and b².

Answer: (a) \( \frac5{\sqrt7} \)

Explanation:
1. The ellipse is \( \frac{x^2}4+\frac{y^2}3=1 \). With m = 1, \( c=-\sqrt7 \) (indeed \( c^2=4+3=7 \)).
2. The point of contact is \( \left(-\frac{a^2m}{c},\frac{b^2}{c}\right)=\left(\frac4{\sqrt7},-\frac3{\sqrt7}\right) \).
3. Its distance from the origin is \( \sqrt{\frac{16+9}{7}}=\frac5{\sqrt7} \).

Teacher's Note:
1. Contact point of y = mx + c with x²/a² + y²/b² = 1 is (−a²m/c, b²/c).
2. Check that the point lies on the line: −3/√7 = 4/√7 − √7.

Answer: (b) \( 45^\circ \)

Explanation:
1. The tangent at the point with eccentric angle \( \theta \) is \( \frac{x}{a}\cos\theta+\frac{y}{b}\sin\theta=1 \).
2. Divide the given line by \( \sqrt2 \): \( \frac{x}{a}\cdot\frac1{\sqrt2}+\frac yb\cdot\frac1{\sqrt2}=1 \).
3. So \( \cos\theta=\sin\theta=\frac1{\sqrt2} \) and \( \theta=45^\circ \).

Teacher's Note:
1. Write the given line in the form (x/a)cos θ + (y/b)sin θ = 1.
2. cos θ = sin θ = 1/√2.

Answer: (a) \( y=0 \)

Explanation:
1. Let P = (a cos θ, b sin θ) and Q = (a cos θ, a sin θ).
2. Tangent at P: \( \frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1 \). Tangent at Q: \( x\cos\theta+y\sin\theta=a \).
3. Multiply the first by a and subtract: \( y\sin\theta\left(\frac ab-1\right)=0 \), so y = 0 (the meeting point is on the major axis).

Teacher's Note:
1. Write both tangents using the same angle θ and subtract.
2. Since a ≠ b, the only possibility is y = 0.

Answer: (c) zero

Explanation:
1. The hyperbola is \( \frac{x^2}{9}-\frac{y^2}{4}=1 \). The given line has slope \( -\frac52 \), so the tangents have slope \( m=\frac25 \).
2. A tangent with slope m exists only if \( c^2=a^2m^2-b^2\ge0 \) (and not equal, for real tangents).
3. Here \( 9\cdot\frac{4}{25}-4=-\frac{64}{25}<0 \), so no tangent exists.

Teacher's Note:
1. The tangent condition for a hyperbola needs c² = a²m² − b² > 0.
2. The slope m must satisfy |m| > b/a.

Answer: (b) \( y=-x\pm3 \)

Explanation:
1. The hyperbola is \( \frac{x^2}{18}-\frac{y^2}{9}=1 \). Perpendicular to y = x means slope −1.
2. c² = a²m² − b² = 18 − 9 = 9, so c = ±3.
3. Tangents: \( y=-x\pm3 \).

Teacher's Note:
1. Use c² = a²m² − b² with m = −1.
2. a² = 18, b² = 9.

Answer: (d) \( 4x+5y=13 \)

Explanation:
1. Use T = S₁: \( 2(2)x+5(1)y-20=2(2)^2+5(1)^2-20 \).
2. The right side is 8 + 5 − 20 = −7, so \( 4x+5y-20=-7 \).
3. Hence \( 4x+5y=13 \).

Teacher's Note:
1. The chord bisected at (h, k) has the equation T = S₁.
2. Be careful to subtract 20 on both sides.

Answer: (a) \( (\sqrt5,3) \)

Explanation:
1. The ellipse is \( \frac{x^2}4+\frac{y^2}5=1 \). Its director circle is \( x^2+y^2=a^2+b^2=9 \).
2. Tangents from a point are perpendicular exactly when the point lies on the director circle.
3. Check: 1 + 8 = 9, 8 + 1 = 9, 4 + 5 = 9, but 5 + 9 = 14 ≠ 9.

Teacher's Note:
1. Perpendicular tangents come from points on the director circle x² + y² = a² + b².
2. Test each point.

Answer: (b) \( x^2+y^2-6y-7=0 \)

Explanation:
1. Here \( a^2=16,\ b^2=9 \), so \( c=\sqrt7 \) and the foci are \( (\pm\sqrt7,0) \).
2. The radius squared is the squared distance from (0, 3) to a focus: \( 7+9=16 \).
3. Circle: \( x^2+(y-3)^2=16 \), i.e. \( x^2+y^2-6y-7=0 \).

Teacher's Note:
1. Find the foci with c² = a² − b².
2. Radius² = (distance from centre to a focus)².

Answer: (b) 27

Explanation:
1. Here \( a=3,\ b^2=5 \), so \( ae=2 \) and the ends of the latera recta are \( \left(\pm2,\pm\frac53\right) \).
2. The tangent at \( \left(2,\frac53\right) \) is \( \frac{2x}9+\frac y3=1 \), with intercepts \( \frac92 \) and 3.
3. By symmetry the quadrilateral is a rhombus with diagonals 9 and 6, so its area is \( \frac12\cdot9\cdot6=27 \).

Teacher's Note:
1. The four tangents form a rhombus symmetric about both axes.
2. Area of a rhombus = ½ × product of diagonals.