JEE Mathematics Circle and Conic Section MCQs Set 25

Answer: (b) \( \left(\dfrac34,\dfrac74\right) \)

Explanation:
1. Solving \( x^2+4x=5 \) gives \( x=1 \), \( y=2 \). The tangent at \( (1,2) \) is \( 2y=2(x+1) \), i.e. \( x-y+1=0 \).
2. Only \( \left(\dfrac34,\dfrac74\right) \) satisfies it. Answer: (b)

Teacher's Note:
1. The tangent at \( (x_1,y_1) \) is \( yy_1=2(x+x_1) \).

Answer: (a) \( \sqrt2+1 \)

Explanation:
1. \( \alpha=\beta^2 \) and the tangent is \( y=\dfrac x{2\beta}+\dfrac\beta2 \). Tangency to the ellipse needs \( c^2=a^2m^2+b^2 \): \( \dfrac{\beta^2}4=\dfrac1{4\beta^2}+\dfrac12 \).
2. So \( \beta^4-2\beta^2-1=0 \), \( \beta^2=1+\sqrt2 \), and \( \alpha=\beta^2=1+\sqrt2 \). Answer: (a)

Teacher's Note:
1. Apply the tangency condition \( c^2=a^2m^2+b^2 \).

Answer: (a) \( 8\pi(3-2\sqrt2) \)

Explanation:
1. The tangent at \( (1,2) \) is \( y=x+1 \), so the centre lies on the normal \( (1+u,2-u) \). Touching the \( x\text{-axis} \): radius \( =|2-u| \) equals \( |u|\sqrt2 \), so \( u^2+4u-4=0 \).
2. \( u=-2\pm2\sqrt2 \), giving radii \( 4\mp2\sqrt2 \). The smaller radius is \( 4-2\sqrt2 \), so the area is \( \pi(24-16\sqrt2)=8\pi(3-2\sqrt2) \). Answer: (a)

Teacher's Note:
1. The centre lies on the normal at the point of contact.

Answer: (c) 5 : 4

Explanation:
1. A tangent \( y=mx+\dfrac3m \) to the parabola is tangent to \( x^2-\dfrac{y^2}8=1 \) when \( \dfrac9{m^2}=m^2-8 \), so \( m^4-8m^2-9=0 \) and \( m=\pm3 \). The common tangents \( y=3x+1 \) and \( y=-3x-1 \) meet at \( P\left(-\dfrac13,0\right) \).
2. The foci are \( S(3,0) \), \( S'(-3,0) \): \( PS=\dfrac{10}3 \) and \( PS'=\dfrac83 \), so \( PS:PS'=5:4 \). Answer: (c)

Teacher's Note:
1. Equate the two tangent conditions for the same slope.

Answer: (c) \( \sqrt3y=x+3 \)

Explanation:
1. A tangent to the parabola is \( y=mx+\dfrac1m \), i.e. \( m^2x-my+1=0 \). Its distance from the centre \( (3,0) \) equals the radius \( 3 \): \( \dfrac{3m^2+1}{\sqrt{m^4+m^2}}=3 \).
2. \( (3m^2+1)^2=9m^2(m^2+1) \) gives \( 3m^2=1 \), \( m=\pm\dfrac1{\sqrt3} \), and one tangent is \( y=\dfrac x{\sqrt3}+\sqrt3 \), i.e. \( \sqrt3y=x+3 \). Answer: (c)

Teacher's Note:
1. Use the tangent form for the parabola and the distance condition for the circle.

Answer: (a) \( x+2y+4=0 \)

Explanation:
1. A tangent to the parabola is \( y=mx+\dfrac1m \). Substituting in \( xy=2 \): \( mx^2+\dfrac xm-2=0 \) must have equal roots, so \( \dfrac1{m^2}+8m=0 \), \( m^3=-\dfrac18 \) and \( m=-\dfrac12 \).
2. The tangent is \( y=-\dfrac x2-2 \), i.e. \( x+2y+4=0 \). Answer: (a)

Teacher's Note:
1. Use the discriminant for tangency to the hyperbola.

Answer: (b) \( 2 \)

Explanation:
1. The tangent \( x-2y+16=0 \) meets the axis at \( A(-16,0) \); the normal \( 2x+y-48=0 \) meets it at \( B(24,0) \). Since \( \angle APB=90^\circ \), \( AB \) is a diameter and \( C=(4,0) \).
2. Slopes: \( PC=\dfrac{16}{12}=\dfrac43 \), \( PB=\dfrac{16}{-8}=-2 \), so \( \tan\theta=\left|\dfrac{\frac43+2}{1-\frac83}\right|=2 \). Answer: (b)

Teacher's Note:
1. The tangent and normal at \( P \) are perpendicular, so \( AB \) is a diameter.

Answer: (a) \( x^2+y^2-4x+8y+12=0 \)

Explanation:
1. \( C=(0,-6) \). For \( P(2t^2,4t) \), \( D^2=4t^4+16t^2+48t+36 \); \( \dfrac{d}{dt}=0 \) gives \( t^3+2t+3=0 \), so \( t=-1 \) and \( P=(2,-4) \).
2. \( CP=\sqrt{4+4}=2\sqrt2 \), so the circle is \( (x-2)^2+(y+4)^2=8 \), i.e. \( x^2+y^2-4x+8y+12=0 \). Answer: (a)

Teacher's Note:
1. Minimise the squared distance using the parametric form.

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

Explanation:
1. \( e=\sqrt{1-\dfrac9{16}}=\dfrac{\sqrt7}4 \), so the foci are \( (\pm\sqrt7,0) \). The radius is the distance from \( (0,3) \): \( \sqrt{7+9}=4 \).
2. The circle is \( x^2+(y-3)^2=16 \), i.e. \( x^2+y^2-6y-7=0 \). Answer: (a)

Teacher's Note:
1. Find the foci, then the distance to the given centre.

Answer: (c) A is true, but R is false.

Explanation:
1. A tangent to the parabola is \( y=mx+\dfrac{\sqrt5}m \). Its distance from the centre \( (0,0) \) must equal the radius \( \sqrt{5/2} \): \( \dfrac{\sqrt5/|m|}{\sqrt{1+m^2}}=\sqrt{\dfrac52} \), so \( m^2(1+m^2)=2 \), i.e. \( m^4+m^2-2=0 \) and \( m=\pm1 \).
2. So the line \( y=x+\sqrt5 \) is a common tangent and the Assertion is true. The condition in the Reason is \( m^4+m^2-2=0 \), not \( m^4-3m^2+2=0 \) (which also allows \( m^2=2 \), giving no common tangent), so the Reason is false. Answer: (c)

Teacher's Note:
1. Derive the condition on \( m \) yourself.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The ellipse is \( \dfrac{x^2}2+\dfrac{y^2}4=1 \); tangency of \( y=mx+\dfrac{4\sqrt3}m \) needs \( \dfrac{48}{m^2}=2m^2+4 \), i.e. \( m^4+2m^2-24=0 \), so \( m^2=4 \) and \( m=\pm2 \).
2. With \( m=2 \) the tangent is \( y=2x+2\sqrt3 \), so the Assertion is true, and the Reason is true. The Reason is only an intermediate condition and does not by itself explain the particular line stated. Answer: (b)

Teacher's Note:
1. Apply the tangency condition for the ellipse.

Answer: (d) \( x^2+4y^2=16 \)

Explanation:
1. The diameter of the first circle is \( 2 \), so \( b=2 \); the diameter of the second is \( 4 \), so \( a=4 \).
2. The ellipse is \( \dfrac{x^2}{16}+\dfrac{y^2}4=1 \), i.e. \( x^2+4y^2=16 \). Answer: (d)

Teacher's Note:
1. Here the diameters are used as the full semi-axis lengths.

Answer: (c) 7

Explanation:
1. The hyperbola is \( \dfrac{x^2}{(12/5)^2}-\dfrac{y^2}{(9/5)^2}=1 \) with \( e^2=1+\dfrac9{16}=\dfrac{25}{16} \), so its foci are \( \left(\pm\dfrac{12}5\cdot\dfrac54,0\right)=(\pm3,0) \).
2. For the ellipse \( ae=3 \), so \( b^2=a^2-(ae)^2=16-9=7 \). Answer: (c)

Teacher's Note:
1. Find the foci of the hyperbola first.

Answer: (a) 4 units

Explanation:
1. \( e=\sqrt{1-\dfrac9{16}}=\dfrac{\sqrt7}4 \), so the foci are \( (\pm\sqrt7,0) \).
2. The distance from \( (0,3) \) to a focus is \( \sqrt{7+9}=4 \). Answer: (a)

Teacher's Note:
1. The radius is the distance from the centre to a focus.