JEE Mathematics Ellipse MCQs Set 07

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

Explanation:
1. Focus-directrix property: \( (x-1)^2 + (y+1)^2 = \frac{1}{4}\cdot\frac{(x - y - 3)^2}{2} \).
2. Multiply by 8: \( 8x^2 + 8y^2 - 16x + 16y + 16 = x^2 + y^2 + 9 - 2xy - 6x + 6y \).
3. Simplify: \( 7x^2 + 2xy + 7y^2 - 10x + 10y + 7 = 0 \).

Teacher's Note:
1. Use distance from focus \( = e \times \) distance from directrix.
2. Divide the directrix expression by \( \sqrt{1^2 + 1^2} \).

Answer: (d) \( \frac{\sqrt{5}}{3} \)

Explanation:
1. \( 4(x+1)^2 + 9(y+2)^2 = 4 + 36 - 4 = 36 \), so \( \frac{(x+1)^2}{9} + \frac{(y+2)^2}{4} = 1 \).
2. \( a^2 = 9,\ b^2 = 4 \).
3. \( e = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3} \).

Teacher's Note:
1. Complete the square in both variables.
2. The larger denominator is \( a^2 \).

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

Explanation:
1. Rewrite as \( \frac{x^2}{r - 2} + \frac{y^2}{5 - r} = 1 \).
2. Both denominators must be positive: \( 2 < r < 5 \).
3. They must also be unequal, otherwise it is a circle: \( r \ne 3.5 \).

Teacher's Note:
1. Remember the ellipse needs two unequal positive denominators.
2. Equal denominators give a circle.

Answer: (c) \( \frac{4}{9} \)

Explanation:
1. \( \frac{x^2}{1/9} + \frac{y^2}{1/4} = 1 \): the major axis is along the y-axis with \( a^2 = \frac{1}{4} \), \( b^2 = \frac{1}{9} \).
2. Latus rectum \( = \frac{2b^2}{a} = \frac{2/9}{1/2} \).
3. So it equals \( \frac{4}{9} \).

Teacher's Note:
1. The larger denominator is \( a^2 \), whichever variable it sits under.
2. The formula is \( \frac{2b^2}{a} \) with \( a > b \).

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

Explanation:
1. The major axis is horizontal and \( ae = 6 - 1 = 5 \), so \( b^2 = a^2 - 25 \).
2. Passing through \( (4, 6) \): \( \frac{9}{a^2} + \frac{16}{a^2 - 25} = 1 \), i.e. \( (a^2 - 5)(a^2 - 45) = 0 \).
3. \( a^2 > 25 \) forces \( a^2 = 45 \) and \( b^2 = 20 \).

Teacher's Note:
1. Distance from centre to focus is \( ae \).
2. Reject the root that makes \( b^2 \) negative.

Answer: (c) on the major axis

Explanation:
1. \( 4(x-2)^2 + 9(y-3)^2 = 36 \), i.e. \( \frac{(x-2)^2}{9} + \frac{(y-3)^2}{4} = 1 \).
2. Centre \( (2, 3) \); major axis is the line \( y = 3 \) and minor axis is \( x = 2 \).
3. The point \( (1, 3) \) has \( y = 3 \), so it lies on the major axis (inside the ellipse as well, since \( \frac{1}{9} < 1 \)).

Teacher's Note:
1. Find the centre and the equations of the two axes.
2. Check which axis equation the point satisfies.

Answer: (b) \( x^2 + y^2 = \frac{7 + 3\sqrt{5}}{2} \)

Explanation:
1. The centre is \( (0, 0) \). The squared semi-axes are \( \frac{1}{\lambda} \) where \( \lambda \) are eigenvalues of \( \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix} \).
2. \( \lambda^2 - 7\lambda + 1 = 0 \), so \( \lambda = \frac{7 \pm 3\sqrt{5}}{2} \).
3. The major semi-axis squared is \( \frac{2}{7 - 3\sqrt{5}} = \frac{7 + 3\sqrt{5}}{2} \), so the auxiliary circle is \( x^2 + y^2 = \frac{7 + 3\sqrt{5}}{2} \).

Teacher's Note:
1. The auxiliary circle has the same centre and radius equal to the semi-major axis.
2. Rationalise \( \frac{2}{7 - 3\sqrt{5}} \) by multiplying by \( 7 + 3\sqrt{5} \).

Answer: (a) an ellipse

Explanation:
1. \( \frac{x}{3} = \cos t + \sin t \) and \( \frac{y}{4} = \cos t - \sin t \).
2. Squaring and adding: \( \frac{x^2}{9} + \frac{y^2}{16} = 2 \).
3. This is \( \frac{x^2}{18} + \frac{y^2}{32} = 1 \), an ellipse.

Teacher's Note:
1. Square and add to eliminate \( t \).
2. Unequal positive coefficients give an ellipse.

Answer: (c) TFTT

Explanation:
1. By definition the latus rectum is the chord through a focus perpendicular to the major axis: S1 true.
2. It does not pass through the centre (it is at distance \( ae \) from it): S2 false.
3. Perpendicular to the major axis means parallel to the minor axis: S3 and S4 true. The sequence is TFTT.

Teacher's Note:
1. Recall the definition of latus rectum.
2. A line perpendicular to the major axis is parallel to the minor axis.

Answer: (b) \( 0 < a < 1 \)

Explanation:
1. Divide by \( 169 \): \( \left(x - \frac{1}{13}\right)^2 + \left(y - \frac{2}{13}\right)^2 = a\left(\frac{5x + 12y - 1}{13}\right)^2 \).
2. This is \( PS^2 = e^2 PM^2 \) with focus \( \left(\frac{1}{13}, \frac{2}{13}\right) \), directrix \( 5x + 12y - 1 = 0 \) and \( e^2 = a \).
3. An ellipse needs \( 0 < e < 1 \), so \( 0 < a < 1 \).

Teacher's Note:
1. Spot the focus-directrix form: \( 5^2 + 12^2 = 13^2 \).
2. \( e < 1 \) characterises an ellipse.

Answer: (b) \( \pm 6 \)

Explanation:
1. Tangent condition: \( c^2 = a^2m^2 + b^2 \).
2. \( c^2 = 8 \times 4 + 4 = 36 \).
3. So \( c = \pm 6 \).

Teacher's Note:
1. Use \( c^2 = a^2m^2 + b^2 \) for the line \( y = mx + c \).
2. Take the square root with both signs.

Answer: (c) \( \frac{5}{\sqrt{7}} \)

Explanation:
1. The ellipse is \( \frac{x^2}{4} + \frac{y^2}{3} = 1 \). The tangent at \( (h, k) \) is \( \frac{xh}{4} + \frac{yk}{3} = 1 \).
2. Compare with \( x - y = \sqrt{7} \), i.e. \( \frac{x}{\sqrt{7}} - \frac{y}{\sqrt{7}} = 1 \): \( h = \frac{4}{\sqrt{7}} \), \( k = -\frac{3}{\sqrt{7}} \).
3. Distance \( = \sqrt{\frac{16 + 9}{7}} = \frac{5}{\sqrt{7}} \).

Teacher's Note:
1. Write the tangent at \( (h, k) \) and match coefficients.
2. Check that \( 3h^2 + 4k^2 = 12 \) holds.

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

Explanation:
1. The tangent at eccentric angle \( \theta \) is \( \frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1 \).
2. Write the given line as \( \frac{x}{a}\cdot\frac{1}{\sqrt{2}} + \frac{y}{b}\cdot\frac{1}{\sqrt{2}} = 1 \).
3. So \( \cos\theta = \sin\theta = \frac{1}{\sqrt{2}} \) and \( \theta = 45^\circ \).

Teacher's Note:
1. Divide the given equation by \( \sqrt{2} \) to get the right side as 1.
2. Compare with the standard tangent form.

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

Explanation:
1. Take \( P(a\cos\phi, b\sin\phi) \) and \( Q(a\cos\phi, a\sin\phi) \).
2. Tangents: \( \frac{x}{a}\cos\phi + \frac{y}{b}\sin\phi = 1 \) and \( \frac{x}{a}\cos\phi + \frac{y}{a}\sin\phi = 1 \).
3. Subtracting gives \( y\sin\phi\left(\frac{1}{b} - \frac{1}{a}\right) = 0 \), so \( y = 0 \): they meet on the major axis.

Teacher's Note:
1. Write both tangents using the same angle \( \phi \).
2. Subtract the equations.

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

Explanation:
1. Use \( T = S_1 \) with midpoint \( (2, 1) \): \( 2x(2) + 5y(1) - 20 = 2(4) + 5(1) - 20 \).
2. \( 4x + 5y - 20 = -7 \).
3. So \( 4x + 5y = 13 \).

Teacher's Note:
1. The chord with a given midpoint is \( T = S_1 \).
2. Compute \( S_1 \) by substituting the midpoint in the curve.

Answer: (d) \( (\sqrt{5}, 3) \)

Explanation:
1. The ellipse is \( \frac{x^2}{4} + \frac{y^2}{5} = 1 \), so the director circle is \( x^2 + y^2 = 4 + 5 = 9 \).
2. Tangents from points on this circle are perpendicular.
3. \( (1, 2\sqrt{2}) \): 1 + 8 = 9; \( (2\sqrt{2}, 1) \): 9; \( (2, \sqrt{5}) \): 9; but \( (\sqrt{5}, 3) \): 5 + 9 = 14 ≠ 9.

Teacher's Note:
1. Director circle: \( x^2 + y^2 = a^2 + b^2 \).
2. Test each point in the circle's equation.

Answer: (d) \( p^2(a^2\sec^2\alpha + b^2\text{cosec}^2\alpha) = (a^2 - b^2)^2 \)

Explanation:
1. The normal at \( (a\cos\phi, b\sin\phi) \) is \( ax\sec\phi - by\,\text{cosec}\phi = a^2 - b^2 \).
2. Compare with \( x\cos\alpha + y\sin\alpha = p \): \( \cos\phi = \frac{ap\sec\alpha}{a^2 - b^2} \) and \( \sin\phi = \frac{-bp\,\text{cosec}\alpha}{a^2 - b^2} \).
3. Squaring and adding: \( p^2(a^2\sec^2\alpha + b^2\text{cosec}^2\alpha) = (a^2 - b^2)^2 \).

Teacher's Note:
1. Match the coefficients of the two forms of the normal.
2. Use \( \cos^2\phi + \sin^2\phi = 1 \).

Answer: (b) 3

Explanation:
1. The product of the perpendicular distances from the two foci to any tangent of an ellipse equals \( b^2 \).
2. Here \( b^2 = 3 \).
3. So \( (S_1F_1)(S_2F_2) = 3 \).

Teacher's Note:
1. This is a standard property of the ellipse.
2. The foci distances to a tangent multiply to \( b^2 \).