JEE Mathematics Parabola MCQs Set 10

Answer: (d) \( (-2,2) \)

Explanation:
1. Rewrite: \( y^2-4y=2x \), so \( (y-2)^2=2x+4=2(x+2) \).
2. The vertex is at \( (-2,2) \).

Teacher's Note:
1. Reduce the expression to a standard form before reading off the answer.
2. Verify with a simple check of the final result.

Answer: (c) \( \left(\frac23,0\right) \)

Explanation:
1. With vertices \( (0,0),(1,2),(1,-2) \), the centroid is \( \left(\frac{0+1+1}{3},0\right)=\left(\frac23,0\right) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) \( \sqrt{13}\,a \)

Explanation:
1. The chord is the polar of the pole: \( y(3a)=2a(x+2a) \), i.e. \( 3y=2x+4a \).
2. Substituting \( x=\frac{3y-4a}{2} \) in \( y^2=4ax \): \( y^2-6ay+8a^2=0 \), so \( y=2a,\ 4a \).
3. Points \( (a,2a) \) and \( (4a,4a) \): length \( \sqrt{9a^2+4a^2}=\sqrt{13}\,a \).

Teacher's Note:
1. The chord with a given pole is the polar (chord of contact).
2. Find the intersection points and use the distance formula.

Answer: (d) 2

Explanation:
1. For \( P(h,s) \) the normals satisfy \( m^3+(2-h)m+s=0 \): \( m_1+m_2+m_3=0 \), \( m_1m_2m_3=-s \), \( \sum m_im_j=2-h \).
2. With \( m_1m_2=k \): \( m_3=-\frac sk \) and \( k+m_3(m_1+m_2)=k-m_3^2=2-h \), so \( s^2=k^2(h+k-2) \).
3. For this to be \( y^2=4x \): \( k^2=4 \) and \( k-2=0 \), so \( k=2 \).

Teacher's Note:
1. Use the cubic satisfied by the slopes of the normals.
2. Match the resulting locus with y squared = 4x.

Answer: (d) \( C_1 \) and \( C_2 \) touch each other at exactly two points

Explanation:
1. Substituting \( y^2=4x \): \( x^2-2x+1=0 \), so \( x=1 \) (double root), giving \( (1,\pm2) \).
2. The repeated root means tangency at each of the two points (the circle \( (x-3)^2+y^2=8 \) has radius \( 2\sqrt2 \) and the normals to the parabola at these points pass through its centre \( (3,0) \)).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) 3

Explanation:
1. Take \( y=x+1 \): \( (x+1)^2=4x+1 \) gives \( x^2-2x=0 \), so \( x=2 \) (x = 0 is on an axis), \( P=(2,3) \).
2. Take \( y=x-1 \): \( x^2-6x=0 \), so \( x=6 \), \( P=(6,5) \).
3. So the ordinate is 3 or 5; of the options only 3 is possible.

Teacher's Note:
1. Write y as x plus or minus 1 and substitute in the parabola.
2. Reject points on the axes.

Answer: (a) \( \frac2r \) or \( -\frac2r \)

Explanation:
1. Let \( A(t_1^2,2t_1),\ B(t_2^2,2t_2) \). The centre has \( y=t_1+t_2 \), and tangency to the x-axis gives \( |t_1+t_2|=r \).
2. Slope of \( AB \): \( \frac{2}{t_1+t_2}=\pm\frac2r \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) the first quadrant

Explanation:
1. Put \( u=x-y \) and \( v=x+y \). The equation becomes \( u^2=8(v-4) \), a parabola.
2. So \( v=4+u^2/8 \). Now \( v-|u|=\frac{u^2-8|u|+32}{8}>0 \) because \( u^2-8|u|+32 \) has negative discriminant.
3. Hence \( x+y>|x-y| \), which gives \( x>0 \) and \( y>0 \): every point lies in the first quadrant.

Teacher's Note:
1. A rotated parabola is easiest to study in the variables x - y and x + y.
2. Showing x + y > |x - y| proves both coordinates are positive.

Answer: (b) \( \left(\frac{9}{16},\frac32\right) \)

Explanation:
1. Slope of \( PQ \) = \( \frac{2-1}{1-\frac14}=\frac43 \).
2. Tangent at \( (t^2,2t) \) has slope \( \frac1t \), so \( t=\frac34 \).
3. Hence \( R=\left(\frac{9}{16},\frac32\right) \).

Teacher's Note:
1. Use the parametric form (t squared, 2t).
2. Equal slopes mean parallel lines.

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

Explanation:
1. The midpoint of \( (a,0) \) and \( (at^2,2at) \) is \( \left(\frac{a(1+t^2)}2,at\right) \), so \( y^2=a(2x-a)=2a\left(x-\frac a2\right) \).
2. Vertex \( \left(\frac a2,0\right) \) and \( 4A=2a \), so \( A=\frac a2 \) and the directrix is \( x=\frac a2-\frac a2=0 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) 3

Explanation:
1. The normal at \( (2t,t^2) \) is \( y-t^2=-\frac1t(x-2t) \). Through \( (1,2) \): \( t^3=1 \), so \( t=1 \).
2. The normal is \( x+y=3 \), so the x-intercept is 3.

Teacher's Note:
1. Work from the given condition to find the function or quantity, then substitute.
2. Verify the answer on a simple case.

Answer: (b) \( (2,-2) \)

Explanation:
1. Rewrite: \( x^2-4x=2y \), so \( (x-2)^2=2y+4=2(y+2) \).
2. The vertex is \( (2,-2) \).

Teacher's Note:
1. Complete the square in x.
2. Read off the vertex from (x - h)^2 = 4a(y - k).

Answer: (c) vertex is \( \left(\frac{2a}{3},0\right) \) and focus is \( (a,0) \)

Explanation:
1. With \( P(at^2,2at) \): \( T(-at^2,0) \), \( N(2a+at^2,0) \). Centroid \( \left(\frac{2a+at^2}{3},\frac{2at}{3}\right) \).
2. Eliminating \( t \): \( y^2=\frac{4a}3\left(x-\frac{2a}3\right) \).
3. Vertex \( (\frac{2a}3,0) \), latus rectum \( \frac{4a}3 \), focus \( (a,0) \), directrix \( x=\frac a3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \frac52 \)

Explanation:
1. The points are \( (0,0),(1,2),(1,-2) \); the circumcentre is on the x-axis at \( (c,0) \) with \( c^2=(1-c)^2+4 \), so \( c=\frac52 \).
2. Circumradius \( =\frac52 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( \sqrt3\,y=x+3 \)

Explanation:
1. A tangent to the parabola: \( y=mx+\frac1m \). Its distance from \( (3,0) \) must be 3: \( \frac{(3m^2+1)^2}{m^2}=9(1+m^2) \).
2. This gives \( 6m^2+1=9m^2 \), \( m^2=\frac13 \). Above the axis \( m=\frac1{\sqrt3} \) and the tangent is \( y=\frac{x}{\sqrt3}+\sqrt3 \), i.e. \( \sqrt3y=x+3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) y = x + 2

Explanation:
1. A tangent to \( y^2=8x \) is \( y=mx+\frac2m \). Substituting in \( xy=-1 \): \( mx^2+\frac2mx+1=0 \).
2. Tangency needs \( \frac4{m^2}-4m=0 \), so \( m^3=1 \), \( m=1 \) and the tangent is \( y=x+2 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) Both are true and Statement 2 is a correct explanation of Statement 1

Explanation:
1. \( y=-\frac12(x-1)^2+\frac32 \), a parabola with vertex \( \left(1,\frac32\right) \) and axis \( x=1 \).
2. It is symmetric about its axis, which explains Statement 1.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( y=0 \) and \( y=4(x-1) \)

Explanation:
1. A tangent to \( y=x^2 \) at \( (t,t^2) \) is \( y=2tx-t^2 \).
2. It touches \( y=-(x-2)^2 \) when \( (x-2)^2+2tx-t^2=0 \), i.e. \( x^2+(2t-4)x+4-t^2=0 \), has zero discriminant: \( (2t-4)^2-4(4-t^2)=8t^2-16t=0 \).
3. So \( t=0 \) or \( t=2 \), giving \( y=0 \) and \( y=4x-4 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( \left(\frac52,0\right) \)

Explanation:
1. The vertices are \( (0,0),(1,2),(1,-2) \). Equating distances from \( (c,0) \): \( c^2=(1-c)^2+4 \) gives \( c=\frac52 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) 9

Explanation:
1. The normal at \( (3t^2,6t) \) has slope \( -t \). For \( x+y=k \) the slope is \( -1 \), so \( t=1 \) and the point is \( (3,6) \).
2. Then \( k=3+6=9 \).

Teacher's Note:
1. Use the parametric point (a t squared, 2 a t) with a = 3.
2. The slope of the normal is minus t.

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

Explanation:
1. \( \frac{x^2}4+y^2=1 \) has \( c=\sqrt3 \), so \( P=\left(-\sqrt3,-\tfrac12\right) \) and \( Q=\left(\sqrt3,-\tfrac12\right) \), with \( PQ=2\sqrt3 \).
2. The focus is \( \left(0,-\frac12\right) \) and \( 4a=2\sqrt3 \). Opening upwards: vertex \( \left(0,-\frac{1+\sqrt3}{2}\right) \), equation \( x^2-2\sqrt3y=3+\sqrt3 \).
3. Opening downwards: vertex \( \left(0,\frac{\sqrt3-1}{2}\right) \), equation \( x^2+2\sqrt3y=3-\sqrt3 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 2

Explanation:
1. Normals through \( (3,0) \) have slopes with \( m^3+(2-3)m=0 \), i.e. \( m=0,\pm1 \), at points \( (0,0),(1,-2),(1,2) \).
2. Area \( =\frac12\cdot4\cdot1=2 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

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

Explanation:
1. Vertex \( (1,1) \), focus \( (2,2) \) (on the axis \( y=x \)). The directrix is perpendicular to the axis at distance \( \sqrt2 \) from the vertex on the other side: \( x+y=0 \).
2. Focus-directrix property: \( (x-2)^2+(y-2)^2=\frac{(x+y)^2}2 \).
3. Simplifying: \( (x-y)^2=8(x+y-2) \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( x=\frac32 \)

Explanation:
1. Complete the square: \( (y+2)^2=-4x+2=-4\left(x-\frac12\right) \).
2. Vertex \( \left(\frac12,-2\right) \), opening left with \( a=1 \), so the directrix is \( x=\frac12+1=\frac32 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (a) Both statements are true and Statement 2 explains Statement 1

Explanation:
1. For \( x^2=6y \) the directrix is \( y=-\frac32 \), i.e. \( 2y+3=0 \) (Statement 2 is true).
2. The point \( \left(-1,-\frac32\right) \) lies on the directrix, so the tangents from it are perpendicular (Statement 1 is true).
3. Statement 2 is exactly the reason for Statement 1.

Teacher's Note:
1. Perpendicular tangents to a parabola always meet on its directrix.
2. Check that the point satisfies the directrix equation.