JEE Mathematics Parabola MCQs Set 14

Answer: (d) \( 0\le\alpha\le1 \)

Explanation:
1. A point is not outside the parabola \( y=x^2+x+1 \) (opening upwards) if it is on it or inside it: \( y\ge x^2+x+1 \).
2. \( 2\alpha+1\ge\alpha^2+\alpha+1 \) gives \( \alpha^2-\alpha\le0 \), so \( 0\le\alpha\le1 \).

Teacher's Note:
1. For an upward-opening parabola the inside is above the curve. 'Not outside' includes points on the curve.

Answer: (b) \( \sqrt{\frac ca} \)

Explanation:
1. The square of the tangent length from \( O \) equals the power of \( O \) with respect to the circle \( =OA\cdot OB=\alpha\beta \).
2. For \( ax^2+bx+c=0 \) the product of the roots is \( \alpha\beta=\frac ca \).
3. The tangent length is \( \sqrt{\frac ca} \).

Teacher's Note:
1. The power of a point is the product of the segments of any secant through it. Here the secant is the x-axis.

Answer: (c) \( \frac{64}{75} \)

Explanation:
1. For the ellipse \( c=3 \) and the latus rectum is the chord \( x=3 \) with end points \( \left(3,\pm\frac{16}{5}\right) \).
2. These lie on the parabola: \( \frac{256}{25}=4a\cdot3 \), so \( a=\frac{256}{300}=\frac{64}{75} \).

Teacher's Note:
1. The latus rectum of the ellipse is the vertical chord \( x=ae \). A double ordinate of a parabola is a chord perpendicular to its axis.

Answer: (c) \( 9x-4y+4=0 \) and \( x-4y+36=0 \)

Explanation:
1. A tangent is \( y=mx+\frac{9}{4m} \). Through \( (4,10) \): \( 10=4m+\frac9{4m} \), so \( 16m^2-40m+9=0 \) and \( m=\frac94 \) or \( \frac14 \).
2. \( m=\frac94 \): \( y=\frac94x+1 \), i.e. \( 9x-4y+4=0 \). \( m=\frac14 \): \( y=\frac x4+9 \), i.e. \( x-4y+36=0 \).

Teacher's Note:
1. Substitute the point into the slope form of the tangent to get a quadratic in \( m \).

Answer: (d) \( y^2-6x+2y-2=0 \) or \( y^2+6x+2y-14=0 \)

Explanation:
1. The focus is the mid-point \( (1,-1) \) and the latus rectum \( =6=4a \), so \( a=\frac32 \). The axis is the horizontal line \( y=-1 \).
2. The vertex is at \( \left(1\mp\frac32,-1\right) \): \( \left(-\frac12,-1\right) \) (opening right) or \( \left(\frac52,-1\right) \) (opening left).
3. \( (y+1)^2=6\left(x+\frac12\right) \) gives \( y^2-6x+2y-2=0 \) and \( (y+1)^2=-6\left(x-\frac52\right) \) gives \( y^2+6x+2y-14=0 \).

Teacher's Note:
1. The ends of the latus rectum lie on a line through the focus perpendicular to the axis, so the focus is their mid-point. Two parabolas are possible.

Answer: (d) \( 4 \)

Explanation:
1. A second degree equation \( ax^2+2hxy+by^2+\dots=0 \) is a parabola when \( h^2=ab \) (and it is not degenerate).
2. Here \( a=\lambda \), \( 2h=4 \) so \( h=2 \), \( b=1 \): \( 4=\lambda \).
3. With \( \lambda=4 \) the quadratic part is \( (2x+y)^2 \) and the linear part \( 4x+3y \) is not a multiple of \( 2x+y \), so the conic is a genuine parabola.

Teacher's Note:
1. The condition \( h^2=ab \) makes the quadratic terms a perfect square. Check non-degeneracy by testing the linear terms.

Answer: (c) \( (-2,0) \)

Explanation:
1. Perpendicular tangents to a parabola meet on its directrix. For \( y^2=8x \) the directrix is \( x=-2 \).
2. On \( y=x+2 \) with \( x=-2 \) we get \( y=0 \): the point is \( (-2,0) \).

Teacher's Note:
1. The locus of the intersection of perpendicular tangents is the directrix (the director circle of a parabola).

Answer: (b) \( x^2+y^2-2x-10y+24=0 \)

Explanation:
1. The tangent at \( (2,4) \) is \( 4y=4(x+2) \), i.e. \( x-y+2=0 \).
2. Circles touching it at that point: \( (x-2)^2+(y-4)^2+\lambda(x-y+2)=0 \). Through \( (0,4) \): \( 4-2\lambda=0 \), so \( \lambda=2 \).
3. Expanding: \( x^2+y^2-2x-10y+24=0 \).

Teacher's Note:
1. A circle touching a line at a point is a point circle plus a multiple of the tangent line. Use the extra point to find the multiple.

Answer: (c) \( \left[\frac\pi2,\frac{5\pi}6\right]\cup\left[\pi,\frac{3\pi}2\right] \)

Explanation:
1. The parabola is \( x=2-2y^2 \) (opening left); a point is not outside it if \( x\le2-2y^2 \).
2. \( \sin\alpha\le2-2\cos^2\alpha=2\sin^2\alpha \), so \( \sin\alpha(2\sin\alpha-1)\ge0 \): \( \sin\alpha\le0 \) or \( \sin\alpha\ge\frac12 \).
3. In \( \left[\frac\pi2,\frac{3\pi}2\right] \): \( \sin\alpha\ge\frac12 \) gives \( \left[\frac\pi2,\frac{5\pi}6\right] \) and \( \sin\alpha\le0 \) gives \( \left[\pi,\frac{3\pi}2\right] \).

Teacher's Note:
1. For a parabola opening left, inside means \( x \) is smaller than the curve value. Solve the resulting inequality in \( \sin\alpha \) over the given interval.

Answer: (d) \( x=a \)

Explanation:
1. For \( P(h,k) \), the slopes satisfy \( hm^2-km+a=0 \), so \( m_1m_2=\frac ah \).
2. \( \theta_1+\theta_2=\frac\pi2 \) means \( \tan\theta_1\tan\theta_2=1 \), so \( \frac ah=1 \).
3. The locus is the line \( x=a \).

Teacher's Note:
1. Angles adding up to \( 90^\circ \) means the product of their tangents is \( 1 \). Use the product of the roots of the quadratic in \( m \).

Answer: (b) the parabolas \( y^2=4(x+1) \) for \( x \lt 2 \) and \( y^2=12(3-x) \) for \( x\ge2 \)

Explanation:
1. Let the point be \( (x,y) \): \( \sqrt{x^2+y^2}+|x-2|=4 \).
2. For \( x \lt 2 \): \( \sqrt{x^2+y^2}=2+x \), so \( y^2=4+4x=4(x+1) \). For \( x\ge2 \): \( \sqrt{x^2+y^2}=6-x \), so \( y^2=36-12x=12(3-x) \).
3. Both arcs meet at \( (2,\pm2\sqrt3) \).

Teacher's Note:
1. The absolute value splits the locus into two cases, each giving a parabola arc. Check continuity at \( x=2 \).

Answer: (a) \( x+3y-33=0 \)

Explanation:
1. Slope of the tangent: \( \frac{dy}{dx}=\frac x2=3 \). So the normal has slope \( -\frac13 \).
2. \( y-9=-\frac13(x-6) \), i.e. \( x+3y-33=0 \).

Teacher's Note:
1. The slope of the normal is the negative reciprocal of the slope of the tangent at the point.

Answer: (b) \( c\le-4 \) or \( c\ge4 \)

Explanation:
1. Shift the origin: \( X=x+2 \), so the parabola is \( y^2=8X \) and the line is \( y=mX+(c-2m) \).
2. Tangency: \( c-2m=\frac2m \), so \( c=2m+\frac2m \).
3. For \( m \gt 0 \), \( c\ge2\cdot2=4 \) (A.M.-G.M.); for \( m \lt 0 \), \( c\le-4 \).

Teacher's Note:
1. After shifting the origin the tangency condition is \( c'=\frac{a}{m} \). Then bound \( 2m+\frac2m \) by A.M.-G.M.

Answer: (c) \( 4x^2-4xy+y^2+8x+46y-71=0 \)

Explanation:
1. The vertex is the mid-point of the focus and the foot of the directrix, so the directrix passes through \( (3,3) \) and is perpendicular to the axis (direction \( (1,2) \) from vertex to focus): \( x+2y-9=0 \).
2. Equating \( SP^2 \) with the squared distance to the directrix: \( 5\left[(x-1)^2+(y+1)^2\right]=(x+2y-9)^2 \).
3. Expanding: \( 4x^2-4xy+y^2+8x+46y-71=0 \).

Teacher's Note:
1. Reflect the focus in the vertex to get a point of the directrix. The directrix is perpendicular to the line joining focus and vertex.

Answer: (a) \( 4y-x-28=0 \) at \( (28,14) \)

Explanation:
1. The slope of the given line is \( \frac14 \). For \( y^2=7x \), \( a=\frac74 \) and the tangent is \( y=mx+\frac am=\frac x4+7 \), i.e. \( 4y-x-28=0 \).
2. The point of contact is \( \left(\frac a{m^2},\frac{2a}m\right)=(28,14) \).

Teacher's Note:
1. Use the slope form of the tangent and the contact point \( \left(\frac{a}{m^2},\frac{2a}{m}\right) \).

Answer: (b) \( y=x^2+2 \)

Explanation:
1. Let \( P=(p,p^2) \), \( Q=(q,q^2) \) with \( OP\perp OQ \): \( pq+p^2q^2=0 \), so \( pq=-1 \).
2. \( R=P+Q=(p+q,\ p^2+q^2)=(h,k) \). Then \( h^2=p^2+q^2+2pq=k-2 \).
3. The locus is \( y=x^2+2 \).

Teacher's Note:
1. In a rectangle \( POQR \) the fourth vertex is \( P+Q \) when \( O \) is the origin. Use \( pq=-1 \) from the right angle.

Answer: (c) \( x+y=3 \) (only one normal)

Explanation:
1. The normal at \( (2t,t^2) \) is \( x+ty=2t+t^3 \). Through \( (1,2) \): \( 1+2t=2t+t^3 \), so \( t^3=1 \).
2. The only real root is \( t=1 \), giving the normal \( x+y=3 \).

Teacher's Note:
1. A cubic in the parameter can have one or three real roots. Here \( t^3=1 \) has only one real solution.

Answer: (b) \( y^2=2a(x-4a) \)

Explanation:
1. Let the chord join \( t_1,t_2 \) with \( t_1t_2=-4 \) (right angle at the vertex). The mid-point is \( \left(\frac{a(t_1^2+t_2^2)}{2},a(t_1+t_2)\right) \).
2. \( y^2=a^2(t_1^2+t_2^2+2t_1t_2)=a^2(t_1^2+t_2^2-8)=2ax-8a^2 \).
3. So \( y^2=2a(x-4a) \).

Teacher's Note:
1. The perpendicularity condition at the vertex is \( t_1t_2=-4 \). Use it to eliminate the parameters from the mid-point.

Answer: (a) \( 8\sqrt2 \)

Explanation:
1. Put \( y-3=x+2 \) in the parabola: with \( u=x+2 \), \( u^2=8u \), so \( u=0 \) or \( u=8 \).
2. The points are \( A(-2,3) \) and \( B(6,11) \).
3. \( AB=\sqrt{8^2+8^2}=8\sqrt2 \).

Teacher's Note:
1. The line is parallel to the shifted coordinates, so substitution gives a simple quadratic in \( u=x+2 \).

Answer: (a) \( -2 \)

Explanation:
1. Here \( 4a=8 \), so \( a=2 \) and the vertex is \( (7,0) \) with the axis vertical, opening upwards.
2. Perpendicular tangents meet on the directrix \( y=-a=-2 \).
3. Hence \( \beta=-2 \).

Teacher's Note:
1. For \( x^2=4ay \)-type parabolas the directrix is \( y=-a \) measured from the vertex.

Answer: (a) \( x+2y-3=0 \)

Explanation:
1. Every normal of the circle passes through its centre \( (-3,-4) \), so \( PQ \) is a chord through \( (-3,-4) \).
2. If the tangents at \( P,Q \) meet at \( (h,k) \), then \( PQ \) is the chord of contact \( ky=2(x+h) \). It passes through \( (-3,-4) \): \( -4k=2(h-3) \).
3. So \( h+2k-3=0 \): the locus is \( x+2y-3=0 \).

Teacher's Note:
1. Every normal to a circle passes through the centre. The chord \( PQ \) is the polar of the intersection point of the tangents.

Answer: (d) \( (4a,4a) \)

Explanation:
1. Put \( x=1-y \) in \( y^2=4ax \): \( y^2+4ay-4a=0 \), so \( y_1+y_2=-4a \). With \( y=2at \): \( t_1+t_2=-2 \).
2. The parameters of the three normals through a point sum to zero: \( t_3=2 \).
3. \( D=(at_3^2,2at_3)=(4a,4a) \).

Teacher's Note:
1. Three concurrent normals have parameters adding to zero. The two known parameters come from the chord equation.

Answer: (a) axis \( y=\frac23 \); latus rectum \( \frac{16}{9} \)

Explanation:
1. Complete the square: \( 9\left(y-\frac23\right)^2-4=16x+57 \), so \( \left(y-\frac23\right)^2=\frac{16}{9}\left(x+\frac{61}{16}\right) \).
2. This has the form \( Y^2=4aX \) with \( 4a=\frac{16}{9} \): the latus rectum is \( \frac{16}{9} \) and the axis is \( y=\frac23 \) (vertex \( \left(-\frac{61}{16},\frac23\right) \)).

Teacher's Note:
1. Complete the square in \( y \) to bring the equation to \( Y^2=4aX \). The axis is where \( Y=0 \).