JEE Mathematics Ellipse MCQs Set 06

Answer: (b) 12 sq units

Explanation:
1. Here \( a=5 \), \( b=4 \), so \( e=\frac35 \) and \( SS'=2ae=6 \).
2. Area \( =\frac12\cdot6\cdot h \), where h is the height of P above the major axis. The height is greatest at the end of the minor axis, where \( h=b=4 \).
3. Maximum area \( =\frac12\cdot6\cdot4=12 \) sq units.

Teacher's Note:
1. The base SS' is fixed, so the area depends only on the height of P.

Answer: (b) \( b^2(p+q)=2apq \)

Explanation:
1. For a focal chord, the semi-latus rectum \( l=\frac{b^2}{a} \) is the harmonic mean of the two segments: \( \frac1p+\frac1q=\frac2l \).
2. So \( \frac{p+q}{pq}=\frac{2a}{b^2} \), that is \( b^2(p+q)=2apq \).

Teacher's Note:
1. Semi-latus rectum is the harmonic mean of the focal chord segments.

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

Explanation:
1. \( ae=2 \) and \( e=\frac12 \), so \( a=4 \).
2. \( b^2=a^2(1-e^2)=16\cdot\frac34=12 \), so the equation is \( \frac{x^2}{16}+\frac{y^2}{12}=1 \).

Teacher's Note:
1. Use \( ae \) for the focus and \( b^2=a^2(1-e^2) \).

Answer: (a) \( 10\sqrt2 \)

Explanation:
1. \( 2ae=8 \) gives \( ae=4 \), and \( \frac{2a}{e}=25 \) gives \( \frac ae=\frac{25}{2} \).
2. Multiplying, \( a^2=4\cdot\frac{25}{2}=50 \), so \( a=5\sqrt2 \).
3. Major axis \( =2a=10\sqrt2 \).

Teacher's Note:
1. Multiply \( ae \) by \( a/e \) to get \( a^2 \).

Answer: (c) \( 5\pm\frac45y_1 \)

Explanation:
1. Here \( b^2=25>a^2=9 \), so the major axis is along the y-axis. \( e^2=1-\frac{9}{25}=\frac{16}{25} \), so \( e=\frac45 \).
2. The foci are \( (0,\pm4) \) and the focal distances are \( b\pm ey_1=5\pm\frac45y_1 \).

Teacher's Note:
1. When the major axis is vertical, the focal distances involve y, not x.

Answer: (a) \( 5x^2+9y^2-54y+36=0 \)

Explanation:
1. The centre is the mid-point \( (0,3) \) and \( c=2 \), \( b^2=5 \), so \( a^2=b^2+c^2=9 \).
2. The foci lie on the line \( y=3 \), so the equation is \( \frac{x^2}{9}+\frac{(y-3)^2}{5}=1 \).
3. Multiplying by 45: \( 5x^2+9(y-3)^2=45 \), which is \( 5x^2+9y^2-54y+36=0 \).

Teacher's Note:
1. Find the centre first, then a and b.

Answer: (c) outside

Explanation:
1. \( S_1=\frac{16}{28}+\frac{9}{20}-1=\frac47+\frac9{20}-1=\frac{80+63-140}{140}=\frac{3}{140} \).
2. \( S_1>0 \), so the point lies outside the ellipse.

Teacher's Note:
1. Substitute the point into the left side and compare with 1.

Answer: (a) an ellipse

Explanation:
1. \( \frac x3=\cos t+\sin t \) 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 and add the two parametric equations.

Answer: (b) \( \left(-\frac25,\frac15\right) \)

Explanation:
1. The tangent at \( (x_1,y_1) \) is \( 4x_1x+9y_1y=1 \) with slope \( -\frac{4x_1}{9y_1} \). The line \( 8x=9y \) has slope \( \frac89 \).
2. So \( -\frac{4x_1}{9y_1}=\frac89 \), that is \( x_1=-2y_1 \).
3. Then \( 16y_1^2+9y_1^2=1 \) gives \( y_1=\pm\frac15 \), \( x_1=\mp\frac25 \). Among the options, \( \left(-\frac25,\frac15\right) \) fits.

Teacher's Note:
1. Equate the slope of the tangent at \( (x_1,y_1) \) to the given slope.

Answer: (b) 1

Explanation:
1. The tangent at \( (a\cos\theta,b\sin\theta) \) is \( \frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1 \), so \( h=\frac{a}{\cos\theta} \) and \( k=\frac{b}{\sin\theta} \).
2. Therefore \( \frac{a^2}{h^2}+\frac{b^2}{k^2}=\cos^2\theta+\sin^2\theta=1 \).

Teacher's Note:
1. Write the tangent in parametric form and read off the intercepts.

Answer: (c) \( \tan^{-1}\left(\pm\frac{b}{ae}\right) \)

Explanation:
1. An end of the latus rectum is \( \left(ae,\frac{b^2}{a}\right)=(a\cos\phi,b\sin\phi) \).
2. So \( \cos\phi=e \) and \( \sin\phi=\frac ba \), giving \( \tan\phi=\frac{b}{ae} \).
3. With the four ends, \( \phi=\tan^{-1}\left(\pm\frac{b}{ae}\right) \) (and the corresponding angles in the other quadrants).

Teacher's Note:
1. Compare \( (ae,\,b^2/a) \) with \( (a\cos\phi,\,b\sin\phi) \).

Answer: (d) \( \frac{4}{\sqrt3} \)

Explanation:
1. The equation is \( \frac{x^2}{4}+\frac{y^2}{12}=1 \), so \( a^2=4 \) and \( b^2=12 \) with \( b>a \) (major axis along y).
2. Latus rectum \( =\frac{2a^2}{b}=\frac{8}{2\sqrt3}=\frac{4}{\sqrt3} \).

Teacher's Note:
1. When \( b>a \), the latus rectum is \( 2a^2/b \).

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

Explanation:
1. For a point \( P(x,y) \): \( SP^2=e^2PM^2 \), so \( (x-1)^2+(y+1)^2=\frac14\cdot\frac{(x-y-3)^2}{2} \).
2. Multiplying by 8: \( 8(x^2+y^2-2x+2y+2)=x^2+y^2+9-2xy-6x+6y \).
3. Hence \( 7x^2+2xy+7y^2-10x+10y+7=0 \).

Teacher's Note:
1. Use the focus-directrix definition \( SP=e\cdot PM \).

Answer: (b) \( \frac{39}{4} \)

Explanation:
1. \( 2ae=10 \) and \( e=\frac58 \), so \( a=8 \) and \( ae=5 \).
2. \( b^2=a^2-(ae)^2=64-25=39 \).
3. Latus rectum \( =\frac{2b^2}{a}=\frac{78}{8}=\frac{39}{4} \).

Teacher's Note:
1. Find a from \( ae \) and e, then \( b^2=a^2-(ae)^2 \).

Answer: (b) 18

Explanation:
1. \( a^2=36 \), \( b^2=20 \), so \( e^2=1-\frac{20}{36}=\frac49 \) and \( e=\frac23 \).
2. The directrices are \( x=\pm\frac ae=\pm9 \), so the distance between them is 18.

Teacher's Note:
1. The directrices are \( x=\pm a/e \).

Answer: (b) \( (0,\pm\sqrt{21}),\ \frac{\sqrt{21}}{5},\ \frac85 \)

Explanation:
1. Since \( 25>4 \) the major axis is along the y-axis: \( e^2=1-\frac{4}{25}=\frac{21}{25} \), so \( e=\frac{\sqrt{21}}{5} \).
2. The foci are \( (0,\pm be)=(0,\pm\sqrt{21}) \).
3. Latus rectum \( =\frac{2a^2}{b}=\frac85 \).

Teacher's Note:
1. Decide the major axis first by comparing the denominators.

Answer: (a) \( \frac{x^2}{144}+\frac{y^2}{169}=1 \)

Explanation:
1. The major axis is along the y-axis with \( b=13 \) (semi-major) and \( c=5 \).
2. The semi-minor axis squared is \( 169-25=144 \), so the equation is \( \frac{x^2}{144}+\frac{y^2}{169}=1 \).

Teacher's Note:
1. The larger denominator goes under the variable along the major axis.

Answer: (c) \( \frac{x^2}{52}+\frac{y^2}{13}=1 \)

Explanation:
1. Let the ellipse be \( \frac{x^2}{A}+\frac{y^2}{B}=1 \). Then \( \frac{16}{A}+\frac9B=1 \) and \( \frac{36}{A}+\frac4B=1 \).
2. Solving, \( A=52 \) and \( B=13 \).
3. Check: \( \frac{16}{52}+\frac9{13}=1 \) and \( \frac{36}{52}+\frac4{13}=1 \).

Teacher's Note:
1. Substitute both points and solve for \( 1/A \) and \( 1/B \).

Answer: (d) an ellipse with eccentricity \( \frac{\sqrt3}{2} \)

Explanation:
1. Completing squares: \( 4(x-3)^2+16(y-1)^2=12+36+16=64 \).
2. So \( \frac{(x-3)^2}{16}+\frac{(y-1)^2}{4}=1 \) with \( a^2=16 \), \( b^2=4 \).
3. \( e=\sqrt{1-\frac4{16}}=\frac{\sqrt3}{2} \).

Teacher's Note:
1. Complete the square in x and y.

Answer: (d) P lies inside C but outside E

Explanation:
1. For P(1, 2): \( \frac19+\frac44=\frac{10}{9}>1 \), so P is outside E. Also \( 1+4=5<9 \), so P is inside C.
2. For Q(2, 1): \( \frac49+\frac14<1 \), so Q is inside E, and \( 4+1=5<9 \) so Q is inside C.
3. Only statement (d) is correct.

Teacher's Note:
1. Substitute each point into both equations.

Answer: (b) 12

Explanation:
1. \( 2b=8 \) gives \( b=4 \). \( b^2=a^2(1-e^2) \) gives \( 16=a^2\cdot\frac49 \), so \( a^2=36 \).
2. Major axis \( =2a=12 \).

Teacher's Note:
1. Use \( b^2=a^2(1-e^2) \).

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

Explanation:
1. The point is \( (\sqrt6\cos\theta,\sqrt2\sin\theta) \), so \( 6\cos^2\theta+2\sin^2\theta=4 \).
2. This gives \( 4\cos^2\theta=2 \), so \( \cos^2\theta=\frac12 \) and \( \theta=\frac\pi4,\frac{3\pi}4,\dots \)
3. Among the options only \( \frac\pi4 \) satisfies this; \( \frac{3\pi}2 \), \( \frac{5\pi}3 \), \( \frac{7\pi}6 \) give squared distances 2, 3 and 5.

Teacher's Note:
1. Use the parametric point \( (a\cos\theta,b\sin\theta) \).

Answer: (b) \( |t|\le1 \)

Explanation:
1. Substituting \( x=at^2 \): \( t^4+\frac{y^2}{b^2}=1 \), so \( y^2=b^2(1-t^4) \).
2. Real points need \( 1-t^4\ge0 \), that is \( |t|\le1 \).

Teacher's Note:
1. Real y requires a non-negative right-hand side.

Answer: (c) 2

Explanation:
1. \( S_1=\frac{49}{16}+\frac{36}{25}-1>0 \), so the point is outside the ellipse.
2. Two tangents can be drawn from an external point.

Teacher's Note:
1. Check the position of the point first.

Answer: (b) \( \sqrt{\frac32} \)

Explanation:
1. The ellipse is \( \frac{x^2}{2}+y^2=1 \) with \( a^2=2 \), \( b^2=1 \). Equally inclined tangents have slope \( m=\pm1 \).
2. \( c^2=a^2m^2+b^2=3 \), so the tangents are \( y=\pm x\pm\sqrt3 \).
3. Distance from the origin \( =\frac{\sqrt3}{\sqrt2}=\sqrt{\frac32} \).

Teacher's Note:
1. Use \( c^2=a^2m^2+b^2 \) with \( m=\pm1 \).

Answer: (b) \( \left(-\frac{2}{\sqrt{11}},\frac{1}{\sqrt{11}}\right) \) or \( \left(\frac{2}{\sqrt{11}},-\frac{1}{\sqrt{11}}\right) \)

Explanation:
1. The tangent at \( (x_1,y_1) \) is \( 2x_1x+3y_1y=1 \) with slope \( -\frac{2x_1}{3y_1} \), which must equal \( \frac43 \). So \( y_1=-\frac{x_1}{2} \).
2. Substituting in the ellipse: \( 2x_1^2+\frac34x_1^2=1 \), so \( x_1^2=\frac4{11} \).
3. So \( x_1=\pm\frac{2}{\sqrt{11}} \) and \( y_1=\mp\frac{1}{\sqrt{11}} \), which is option (b).

Teacher's Note:
1. The point has opposite-sign coordinates because the required slope is positive.

Answer: (c) \( x^2+y^2=13 \)

Explanation:
1. Perpendicular tangents meet on the director circle \( x^2+y^2=a^2+b^2 \).
2. Here \( a^2+b^2=9+4=13 \).

Teacher's Note:
1. Director circle: \( x^2+y^2=a^2+b^2 \).

Answer: (a) \( \frac{1}{2x^2}+\frac{1}{4y^2}=1 \)

Explanation:
1. The ellipse is \( \frac{x^2}{2}+y^2=1 \); the tangent \( \frac{x\cos\theta}{\sqrt2}+y\sin\theta=1 \) meets the axes at \( \left(\frac{\sqrt2}{\cos\theta},0\right) \) and \( \left(0,\frac1{\sin\theta}\right) \).
2. The mid-point is \( (h,k)=\left(\frac{1}{\sqrt2\cos\theta},\frac{1}{2\sin\theta}\right) \), so \( \cos\theta=\frac1{\sqrt2h} \) and \( \sin\theta=\frac1{2k} \).
3. Using \( \cos^2\theta+\sin^2\theta=1 \): \( \frac1{2h^2}+\frac1{4k^2}=1 \).

Teacher's Note:
1. Parametrise the tangent and eliminate \( \theta \).

Answer: (a) \( \frac\pi6 \)

Explanation:
1. Divide the line by \( 2ab \): \( \frac{\sqrt3x}{2a}+\frac{y}{2b}=1 \).
2. Compare with the tangent \( \frac{x\cos\phi}{a}+\frac{y\sin\phi}{b}=1 \): \( \cos\phi=\frac{\sqrt3}2 \) and \( \sin\phi=\frac12 \).
3. So \( \phi=\frac\pi6 \).

Teacher's Note:
1. Match coefficients with the parametric tangent.

Answer: (c) 1

Explanation:
1. The known relation is \( p^2=\frac{b^2r}{2a-r} \).
2. So \( \frac{b^2}{p^2}=\frac{2a-r}{r}=\frac{2a}{r}-1 \).
3. Hence \( \frac{2a}{r}-\frac{b^2}{p^2}=1 \).

Teacher's Note:
1. Use the pedal relation \( p^2=b^2r/(2a-r) \).