JEE Mathematics Parabola MCQs Set 13

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

Explanation:
1. Tangent at \( (1,2) \): \( y=x+1 \). The centre lies on the normal \( x+y=3 \), so \( (h,k)=(3-k,k) \) and, touching the x-axis, the radius is \( |k| \).
2. Distance from the centre to the tangent equals \( |k| \): \( \frac{|4-2k|}{\sqrt2}=|k| \), so \( k^2-8k+8=0 \) and \( k=4\pm2\sqrt2 \).
3. The smaller radius is \( 4-2\sqrt2 \): area \( =\pi(4-2\sqrt2)^2=\pi(24-16\sqrt2)=8\pi(3-2\sqrt2) \).

Teacher's Note:
1. The centre of a circle touching a curve at a point lies on the normal there. Equate the distances from the centre to the tangent line and the x-axis.

Answer: (d) (i), (iii) and (iv) only

Explanation:
1. The circle has centre \( S(2,8) \) and radius \( 2 \). The nearest point \( P(t^2,2t) \) has the normal through \( S \): \( 8+2t=2t+t^3 \), so \( t=2 \) and \( P=(4,4) \).
2. \( SP=\sqrt{4+16}=2\sqrt5 \) (true). \( SQ=2 \), \( QP=2\sqrt5-2 \), so \( SQ:QP=1:(\sqrt5-1)=(\sqrt5+1):4 \), so (ii) is false.
3. The normal at \( t=2 \) is \( y+2x=12 \), with x-intercept \( 6 \) (true). \( SP \) has slope \( -2 \), so the tangent at \( Q \) is perpendicular to it with slope \( \frac12 \) (true).

Teacher's Note:
1. The shortest distance from an external point to a curve is along a normal. Statement (ii) fails because \( SQ=2 \) is the radius, not \( (\sqrt5+1) \) in a ratio to \( 2 \).

Answer: (d) \( \frac{t^2-1}{t} \)

Explanation:
1. Since \( PQ \) is a focal chord, \( Q \) has parameter \( -\frac1t \).
2. Slope of \( PK=\frac{2at}{at^2-2a}=\frac{2t}{t^2-2} \). Slope of \( QR=\frac{2}{r-\frac1t} \).
3. Equating: \( r-\frac1t=\frac{t^2-2}{t} \), so \( r=\frac{t^2-1}{t} \).

Teacher's Note:
1. The slope of a chord joining parameters \( p \) and \( q \) on \( y^2=4ax \) is \( \frac{2}{p+q} \). A focal chord has parameters whose product is \( -1 \).

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

Explanation:
1. A tangent of the parabola with slope \( m \) is \( y=mx+\frac1m \). It cuts the circle in a chord only if its distance from the centre is less than 2: \( \frac{1}{|m|\sqrt{1+m^2}} \lt 2 \).
2. So \( 4m^2(1+m^2) \gt 1 \), i.e. \( m^2 \gt \frac{\sqrt2-1}{2} \).
3. Hence \( |m| \gt \sqrt{\frac{\sqrt2-1}{2}}\approx0.455 \). Intervals (i) and (iii) lie entirely in this region; (ii) and (iv) contain values of \( |m| \lt 0.455 \).

Teacher's Note:
1. A line meets a circle in a chord when its distance from the centre is smaller than the radius. Solve the resulting inequality in \( m^2 \), a quadratic in \( m^2 \).

Answer: (b) \( x^2+y^2-13x+2y+12=0 \)

Explanation:
1. \( P=(4,4) \) and the tangent at \( P \) is \( x-2y+4=0 \). The focus is \( (1,0) \).
2. Circle touching the tangent at \( P \): \( (x-4)^2+(y-4)^2+\lambda(x-2y+4)=0 \). Passing through \( (1,0) \): \( 9+16+5\lambda=0 \), so \( \lambda=-5 \).
3. Expanding: \( x^2+y^2-13x+2y+12=0 \).

Teacher's Note:
1. The family of circles touching a line at a given point is (circle of zero radius at the point) plus \( \lambda \) times the line. The condition of passing through the focus fixes \( \lambda \).

Answer: (c) \( \frac{a(t^2+1)^2}{2t^3} \)

Explanation:
1. Tangent at \( P \): \( ty=x+at^2 \). Normal at \( S \): \( y+sx=2as+as^3 \).
2. Substituting \( x=ty-at^2 \): \( y(1+st)=2as+as^3+ast^2 \). With \( s=\frac1t \): \( 2y=\frac{2a}{t}+\frac{a}{t^3}+at \).
3. \( y=\frac{a(t^4+2t^2+1)}{2t^3}=\frac{a(t^2+1)^2}{2t^3} \).

Teacher's Note:
1. Eliminate \( x \) between the tangent at \( P \) and the normal at \( S \), then use \( st=1 \). Notice \( t^4+2t^2+1 \) is a perfect square.

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

Explanation:
1. Tangent to the parabola: \( y=mx+\frac1m \). Put it in \( xy=2 \): \( mx^2+\frac xm-2=0 \).
2. Tangency needs a zero discriminant: \( \frac{1}{m^2}+8m=0 \), so \( m^3=-\frac18 \) and \( m=-\frac12 \).
3. The line is \( y=-\frac x2-2 \), i.e. \( x+2y+4=0 \).

Teacher's Note:
1. Take the parabola tangent in slope form and force it to touch the hyperbola by setting the discriminant of the quadratic in \( x \) to zero.

Answer: (b) \( \frac{20}{7} \)

Explanation:
1. Points are \( (p^2,2p) \) and \( (q^2,2q) \), so the slope of \( PQ \) is \( \frac{2p-2q}{p^2-q^2}=\frac{2}{p+q} \).
2. The line \( 10y=7x \) has slope \( \frac7{10} \), so \( \frac{2}{p+q}=\frac{7}{10} \) and \( p+q=\frac{20}{7} \).

Teacher's Note:
1. For \( y^2=4ax \) the slope of the chord joining parameters \( p \) and \( q \) is always \( \frac{2}{p+q} \). Parallel chords share the same value of \( p+q \).

Answer: (a) The orthocentre of the tangent triangle lies on \( x=\frac a3 \)

Explanation:
1. For any parabola the area of the triangle formed by three tangents is half the area of the triangle of contact points, so \( \Delta_1=2\Delta_2 \). This rules out the first three statements.
2. The orthocentre of a triangle formed by three tangents to a parabola lies on its directrix. The directrix of \( P_1 \) is \( x=\frac{2a}{3}-\frac{a}{3}=\frac a3 \).

Teacher's Note:
1. Two standard results about tangential triangles of a parabola are used: area ratio \( 2:1 \) and orthocentre on the directrix. The latus rectum of \( P_1 \) is \( \frac{4a}{3} \), so its focal parameter is \( \frac a3 \).

Answer: (a) \( \frac{81}{4} \)

Explanation:
1. Tangent: \( y=mx+\frac{9}{4m} \). Its distance from \( O \) is \( OP=\frac{9}{4|m|\sqrt{1+m^2}} \).
2. The perpendicular from \( O \) is \( y=-\frac{x}{m} \), i.e. \( x=-my \). Putting this in \( y^2=9x \) gives \( y=-9m \), \( x=9m^2 \), so \( OQ=9|m|\sqrt{1+m^2} \).
3. Hence \( OP\cdot OQ=\frac{9}{4}\cdot 9=\frac{81}{4} \).

Teacher's Note:
1. Here the perpendicular from the vertex to the tangent gives the foot \( P \), and the product turns out to be independent of the tangent chosen. Check with \( m=1 \) if you want a quick numerical confirmation.

Answer: (a) \( 3 \)

Explanation:
1. Take \( A,B,C \) with parameters \( t_1,t_2,t_3 \). The tangents at \( t_i,t_j \) meet at \( (at_it_j,\,a(t_i+t_j)) \).
2. Computing both determinants gives \( [ABC]=a^2|(t_1-t_2)(t_2-t_3)(t_3-t_1)| \) and \( [PQR]=\frac{a^2}{2}|(t_1-t_2)(t_2-t_3)(t_3-t_1)| \).
3. So the ratio is \( \frac21 \) and \( \lambda+\mu=3 \).

Teacher's Note:
1. The triangle formed by three tangents to a parabola has exactly half the area of the triangle formed by the three points of contact. This result is worth remembering for all parabola problems of this kind.

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

Explanation:
1. Both curves pass through the origin and have the y-axis as common tangent there. The circle has centre \( (-b,0) \).
2. External contact needs the two curves on opposite sides of the tangent line \( x=0 \). For \( a \gt 0 \) the parabola lies to the right, so the centre \( -b \) must be negative: \( b \gt 0 \).
3. For \( a \lt 0 \) the parabola lies to the left, so \( -b \gt 0 \), i.e. \( b \lt 0 \). Hence (i) and (iv).

Teacher's Note:
1. External tangency at the origin means the curves lie on opposite sides of the common tangent. Decide the side of each curve from the signs of \( a \) and the centre.

Answer: (b) \( \frac{1}{2^9} \)

Explanation:
1. The parabola is \( Y^2=4aX \) with \( 4a=\frac18 \), so \( a=\frac1{32} \) and the latus rectum is \( \frac18 \).
2. The triangle has base equal to the latus rectum and height \( a \) (the distance from the vertex to the focus).
3. Area \( =\frac12\cdot\frac18\cdot\frac1{32}=\frac{1}{512}=\frac1{2^9} \).

Teacher's Note:
1. The triangle on the latus rectum with the vertex always has area \( 2a^2 \). Here \( 2a^2=2\cdot\frac1{1024}=\frac1{512} \).

Answer: (b) \( 23 \)

Explanation:
1. The focus is \( (0,1) \). The tangent at \( (6,9) \) is \( 3x-y-9=0 \), so the centre lies on the normal through \( (6,9) \) with direction \( (3,-1) \): centre \( =(6+3k,\,9-k) \).
2. Radius\( ^2 \) is \( 10k^2 \). Equating with the squared distance to the focus: \( (6+3k)^2+(8-k)^2=10k^2 \), which gives \( 100+20k=0 \), so \( k=-5 \).
3. The centre is \( (-9,14) \), so \( |\alpha-\beta|=23 \).

Teacher's Note:
1. A circle touching a curve at a point has its centre on the normal at that point. Use the fact that the focus lies on the circle to fix the position along the normal.

Answer: (d) \( 2 \)

Explanation:
1. At \( P(16,16) \) the tangent is \( 2y=x+16 \), meeting the axis at \( A(-16,0) \). The normal has slope \( -2 \): \( y-16=-2(x-16) \), meeting the axis at \( B(24,0) \).
2. The tangent and normal are perpendicular, so \( \angle APB=90^\circ \) and \( AB \) is a diameter: \( C=(4,0) \), \( CP=CB=20 \).
3. \( \theta=\angle CPB=\angle CBP \), and \( \tan\angle CBP=\frac{16}{24-16}=2 \).

Teacher's Note:
1. A right angle at \( P \) makes \( AB \) the diameter of the circle. Equal radii \( CP=CB \) make the base angles of triangle \( CPB \) equal.

Answer: (a) \( -\frac23\sqrt5 \)

Explanation:
1. As before \( t_1t_2=-1 \) and \( t_1+t_2=-1 \), so \( (t_1-t_2)^2=1+4=5 \).
2. The slopes of \( OP \) and \( OQ \) are \( \frac2{t_1} \) and \( \frac2{t_2} \): \( m_1m_2=\frac{4}{t_1t_2}=-4 \) and \( |m_1-m_2|=\frac{2|t_1-t_2|}{|t_1t_2|}=2\sqrt5 \).
3. \( \tan\theta=\frac{m_1-m_2}{1+m_1m_2} \) has magnitude \( \frac{2\sqrt5}{3} \) and, since \( 1+m_1m_2=-3 \lt 0 \), the angle is obtuse, so \( \tan\theta=-\frac23\sqrt5 \).

Teacher's Note:
1. When \( 1+m_1m_2 \) is negative the angle between the two lines at the vertex is obtuse, so the tangent is negative. Always check this sign for the angle subtended by a focal chord.

Answer: (d) \( 15 \)

Explanation:
1. Tangent \( y=mx+\frac2m \) to the parabola is tangent to the circle if \( \frac{4}{m^2}=2(1+m^2) \), i.e. \( m^4+m^2-2=0 \), so \( m=\pm1 \): the lines \( y=x+2 \) and \( y=-x-2 \).
2. Contact points: circle \( P(-1,1) \), \( Q(-1,-1) \); parabola \( R(2,4) \), \( S(2,-4) \).
3. \( PQRS \) is a trapezium with parallel sides \( 2 \) and \( 8 \) and height \( 3 \): area \( =\frac{(2+8)}{2}\cdot3=15 \).

Teacher's Note:
1. Solve for the common slope first, then locate the four points of contact. The two parallel sides of the quadrilateral are vertical.

Answer: (b) \( \frac{4a}{3} \)

Explanation:
1. Let \( P=(at_1^2,2at_1) \), \( Q=(at_2^2,2at_2) \) with \( t_1t_2=-1 \) for a focal chord. The centroid is \( \left(\frac{a(t_1^2+t_2^2)}{3},\frac{2a(t_1+t_2)}{3}\right) \).
2. \( y^2=\frac{4a^2(t_1^2+t_2^2-2)}{9}=\frac{4a}{3}x-\frac{8a^2}{9}=\frac{4a}{3}\left(x-\frac{2a}{3}\right) \).
3. The latus rectum is \( \frac{4a}{3} \).

Teacher's Note:
1. Use \( t_1t_2=-1 \) to eliminate the parameters. The coefficient of \( x \) in \( y^2=\lambda(x-h) \) is the latus rectum.

Answer: (d) \( \left(\frac{2a}{3},0\right) \)

Explanation:
1. For a focal chord \( t_1t_2=-1 \), so the centroid \( (h,k) \) satisfies \( h=\frac{a(t_1^2+t_2^2)}{3} \), \( k=\frac{2a(t_1+t_2)}{3} \).
2. Then \( k^2=\frac{4a}{3}\left(h-\frac{2a}{3}\right) \).
3. The vertex is \( \left(\frac{2a}{3},0\right) \).

Teacher's Note:
1. The minimum value of \( t_1^2+t_2^2 \) is \( 2 \) (when \( t_1+t_2=0 \)), giving the smallest \( h=\frac{2a}{3} \). That is the vertex of the locus.

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

Explanation:
1. Write \( P\left(\frac{t_1^2}{2},t_1\right) \), \( Q\left(\frac{t_2^2}{2},t_2\right) \). \( OP\perp OQ \) gives \( \frac{t_1^2t_2^2}{4}+t_1t_2=0 \), so \( t_1t_2=-4 \).
2. Area \( =\frac12\left|x_1y_2-x_2y_1\right|=\frac14|t_1t_2||t_1-t_2|=|t_1-t_2|=3\sqrt2 \), so \( (t_1+t_2)^2=18-16=2 \).
3. Then \( (t_1,t_2)=(2\sqrt2,-\sqrt2) \) or \( (\sqrt2,-2\sqrt2) \), giving \( P=(4,2\sqrt2) \) or \( P=(1,\sqrt2) \).

Teacher's Note:
1. The right angle at the vertex gives \( t_1t_2=-4 \). Use the area to find \( |t_1-t_2| \) and then solve for each parameter.

Answer: (c) \( 3 \)

Explanation:
1. A normal of slope \( m \) is \( y=mx-2m-m^3 \). Passing through \( (c,0) \) gives \( m\left(m^2+2-c\right)=0 \).
2. The slopes are \( 0 \) and \( \pm\sqrt{c-2} \). Two perpendicular normals must be the pair \( \pm\sqrt{c-2} \), so \( -(c-2)=-1 \).
3. Hence \( c=3 \).

Teacher's Note:
1. Substitute the point into the slope form of the normal and solve the cubic in \( m \). The slope \( 0 \) is the axis itself, which cannot be perpendicular to a normal with finite non-zero slope.

Answer: (c) (ii) and (iii) only

Explanation:
1. Let the chords be \( OP,OQ \) with \( P=(at_1^2,2at_1) \), \( Q=(at_2^2,2at_2) \), perpendicular so \( t_1t_2=-4 \).
2. The mid-points \( M_1\left(\frac{at_1^2}{2},at_1\right) \), \( M_2\left(\frac{at_2^2}{2},at_2\right) \) also subtend a right angle at \( O \), so \( M_1M_2 \) is a diameter and the centre is the mid-point \( \left(\frac{a(t_1^2+t_2^2)}{4},\frac{a(t_1+t_2)}{2}\right) \).
3. Then \( y^2=\frac{a^2(t_1^2+t_2^2-8)}{4}=ax-2a^2 \), i.e. \( y^2=a(x-2a) \): vertex \( (2a,0) \), latus rectum \( a \).

Teacher's Note:
1. A circle through the vertex and two points that subtend a right angle there has those two points as ends of a diameter. The centre is then their mid-point.

Answer: (b) \( (1,1,3) \)

Explanation:
1. The normal of \( y^2=4b(x-c) \) with slope \( m \) is \( y=m(x-c)-2bm-bm^3 \); for \( y^2=8ax \) it is \( y=mx-4am-2am^3 \).
2. A common normal other than the axis (\( m\neq0 \)) needs \( -mc-2bm-bm^3=-4am-2am^3 \), i.e. \( m^2(b-2a)=4a-c-2b \).
3. So \( m^2=\frac{4a-c-2b}{b-2a} \gt 0 \). For \( (1,1,3) \): \( \frac{4-3-2}{1-2}=1 \gt 0 \). The other triads give \( -5,-2,-2 \lt 0 \).

Teacher's Note:
1. A common normal needs a real value of the slope, so \( m^2 \) must be positive. Test each triad in \( m^2=\frac{4a-c-2b}{b-2a} \).

Answer: (c) \( -128 \)

Explanation:
1. Tangent to \( y^2=4x \): \( c=\frac1m \), so \( 4=\frac1m \) and \( m=\frac14 \).
2. For \( x^2=2by \): substitute \( y=mx+4 \) to get \( x^2-2bmx-8b=0 \). Tangency: \( 4b^2m^2+32b=0 \), so \( bm^2=-8 \).
3. With \( m^2=\frac1{16} \): \( b=-128 \).

Teacher's Note:
1. Use the condition \( c=\frac am \) for the first parabola and a zero discriminant for the second.

Answer: (c) \( 95 \)

Explanation:
1. Differentiating \( y=x^2+6 \) gives slope \( 2 \) at \( (1,7) \). The tangent is \( y-7=2(x-1) \), i.e. \( 2x-y+5=0 \).
2. The circle has centre \( (-8,-6) \) and \( r^2=100-c \). Distance from the centre to the line: \( \frac{|-16+6+5|}{\sqrt5}=\sqrt5 \).
3. So \( 100-c=5 \) and \( c=95 \).

Teacher's Note:
1. Tangency of a line to a circle means distance from centre equals the radius. Keep track of \( r^2=g^2+f^2-c \).

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

Explanation:
1. Tangent \( y=mx+\frac1m \) to the parabola touches the circle when \( \frac{1}{m^2(1+m^2)}=\frac12 \), so \( m^2=1 \): the lines \( y=x+1 \) and \( y=-x-1 \) meet at \( Q(-1,0) \), so \( OQ=1 \).
2. The ellipse is \( x^2+\frac{y^2}{1/2}=1 \): \( e^2=1-\frac12 \), so \( e=\frac1{\sqrt2} \), and the latus rectum is \( \frac{2b^2}{a}=1 \).
3. Area between \( x=\frac1{\sqrt2} \) and \( x=1 \) is \( 2\int\frac{1}{\sqrt2}^{1}\sqrt{\frac{1-x^2}{2}}\,dx=\sqrt2\left(\frac\pi8-\frac14\right)=\frac{\pi-2}{4\sqrt2} \). So (i) and (iii).

Teacher's Note:
1. Use \( \int\sqrt{1-x^2}dx=\frac12\left(x\sqrt{1-x^2}+\sin^{-1}x\right) \) to get the area. The two common tangents are symmetric about the x-axis and meet on it.