JEE Mathematics Parabola MCQs Set 17

Answer: (c) \( \dfrac7{4\sqrt2} \)

Explanation:
1. The nearest point of the parabola has tangent slope 1: \( 2y\,y'=1 \) gives \( y=\frac12 \), \( x=\frac94 \), the point \( P\left(\frac94,\frac12\right) \).
2. Distance from \( P \) to \( x-y=0 \) is \( \dfrac{|9/4-1/2|}{\sqrt2}=\dfrac7{4\sqrt2} \). Answer: (c)

Teacher's Note:
1. The shortest distance is along the common normal, where the tangent is parallel to the line.

Answer: (b) 25

Explanation:
1. The focus is \( (4,0) \), so the chord has slope \( \tan\alpha=\dfrac{4-0}{1-4}=-\dfrac43 \) and \( \cot\alpha=-\dfrac34 \).
2. Length \( =4a\,\mathrm{cosec}^2\alpha=16(1+\cot^2\alpha)=16\cdot\dfrac{25}{16}=25 \). Answer: (b)

Teacher's Note:
1. A focal chord inclined at \( \alpha \) has length \( 4a\,\mathrm{cosec}^2\alpha \).

Answer: (c) \( \sqrt2 \)

Explanation:
1. Tangent to the parabola with slope \( -a \): \( y=-ax-\dfrac{\sqrt2}a \), so \( c=-\dfrac{\sqrt2}a \). Tangent to the circle: \( \dfrac{|c|}{\sqrt{1+a^2}}=1 \).
2. So \( 1+a^2=\dfrac2{a^2} \), \( a^4+a^2-2=0 \), \( a^2=1 \), and \( |c|=\dfrac{\sqrt2}{|a|}=\sqrt2 \). Answer: (c)

Teacher's Note:
1. Use the tangent condition for both curves.

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

Explanation:
1. A tangent to the parabola is \( y=mx+\dfrac4m \). Substituting in \( xy=-4 \): \( mx^2+\dfrac4mx+4=0 \) must have equal roots.
2. Discriminant: \( \dfrac{16}{m^2}-16m=0 \), so \( m^3=1 \), \( m=1 \) and the tangent is \( y=x+4 \). Answer: (a)

Teacher's Note:
1. A tangent to the hyperbola meets it at one point only.

Answer: (c) \( (8,6) \)

Explanation:
1. The vertex is \( (2,0) \) and \( a=4-2=2 \), so the parabola is \( y^2=8(x-2) \).
2. \( (4,-4) \): \( 16=16 \); \( (6,4\sqrt2) \): \( 32=32 \); \( (5,2\sqrt6) \): \( 24=24 \); but \( (8,6) \): \( 36\ne48 \). Answer: (c)

Teacher's Note:
1. Shift the origin to the vertex.

Answer: (a) \( 31\dfrac14 \)

Explanation:
1. Take \( C=(t^2,2t) \) with \( A \) at \( t=-2 \) and \( B \) at \( t=3 \). The area is \( \dfrac12\left|\det\right|=5|t^2-t-6| \).
2. On the arc \( -2\lt t\lt3 \) the expression is negative with greatest magnitude at \( t=\dfrac12 \): \( 5\left(6+\dfrac12-\dfrac14\right)=\dfrac{125}4=31\dfrac14 \). Answer: (a)

Teacher's Note:
1. Parametrise points of the parabola as \( (t^2,2t) \).

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

Explanation:
1. Normals of slope \( m \): \( y=m(x-c)-2bm-bm^3 \) and \( y=mx-4am-2am^3 \). Equating gives \( m\left[(2a-b)m^2+(4a-2b-c)\right]=0 \).
2. For \( m\ne0 \): \( m^2=\dfrac c{2a-b}-2\gt0 \), so \( \dfrac c{2a-b}\gt2 \). This holds only for \( (1,1,3) \) (it gives \( 3\gt2 \)). Answer: (c)

Teacher's Note:
1. A non-axis common normal needs a positive value of \( m^2 \).

Answer: (d) \( 6\sqrt3 \)

Explanation:
1. Substituting \( y=\dfrac{x+4\sqrt2}{\sqrt2} \) in \( x^2=4y \): \( x^2-2\sqrt2x-16=0 \), so \( |x_1-x_2|=\sqrt{8+64}=6\sqrt2 \).
2. The slope is \( \dfrac1{\sqrt2} \), so the chord length is \( 6\sqrt2\cdot\sqrt{1+\dfrac12}=6\sqrt3 \). Answer: (d)

Teacher's Note:
1. Chord length \( =|x_1-x_2|\sqrt{1+m^2} \).

Answer: (b) \( 5 \)

Explanation:
1. The vertex is \( (a^2,0) \); at \( x=0 \), \( y^2=4a^2 \), so the other vertices are \( (0,\pm2a) \).
2. Area \( =\dfrac12(4a)(a^2)=2a^3=250 \), so \( a=5 \). Answer: (b)

Teacher's Note:
1. The base lies on the y-axis with length \( 4a \).

Answer: (d) \( \dfrac{125}4 \)

Explanation:
1. With \( X=(t^2,2t) \), \( t\in(-2,3) \), the area is \( 5|t^2-t-6|=-5(t^2-t-6) \).
2. The maximum of \( -5t^2+5t+30 \) is at \( t=\dfrac12 \): \( \dfrac{125}4 \). Answer: (d)

Teacher's Note:
1. Use the determinant formula for the area.

Answer: (c) 32

Explanation:
1. The rectangle with vertices \( (\pm a,0) \) and \( (\pm a,12-a^2) \) has area \( A=2a(12-a^2)=24a-2a^3 \).
2. \( A'=24-6a^2=0 \) gives \( a=2 \), and \( A=48-16=32 \). Answer: (c)

Teacher's Note:
1. Use symmetry about the y-axis and maximise with calculus.

Answer: (d) \( x^2=2y \)

Explanation:
1. Take \( Q=(4t,2t^2) \). The point dividing \( OQ \) in the ratio \( 1:3 \) is \( P=\left(t,\dfrac{t^2}2\right) \).
2. Eliminating \( t \): \( y=\dfrac{x^2}2 \), i.e. \( x^2=2y \). Answer: (d)

Teacher's Note:
1. Use the section formula with a parametric point.

Answer: (a) \( \dfrac12 \)

Explanation:
1. A tangent to \( y^2=4x \) is \( y=mx+\dfrac1m \). Substituting into \( x^2=-32y \): \( x^2+32mx+\dfrac{32}m=0 \).
2. Tangency needs a zero discriminant: \( 1024m^2=\dfrac{128}m \), so \( m^3=\dfrac18 \) and \( m=\dfrac12 \). Answer: (a)

Teacher's Note:
1. A tangent meets the other curve in coincident points.

Answer: (c) \( x=-1 \)

Explanation:
1. Perpendicular tangents to a parabola meet on its directrix.
2. The directrix of \( y^2=4x \) is \( x=-1 \). Answer: (c)

Teacher's Note:
1. Use the directrix property.

Answer: (c) \( (1,0) \)

Explanation:
1. The vertex is the mid-point of the focus and the foot of the directrix on the axis.
2. The foot of the directrix is \( (2,0) \) and the focus is \( (0,0) \), so the vertex is \( (1,0) \). Answer: (c)

Teacher's Note:
1. The axis is perpendicular to the directrix through the focus.

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

Explanation:
1. Perpendicular tangents intersect on the directrix, which is \( x=-2 \).
2. On \( y=x+2 \) the point with \( x=-2 \) is \( (-2,0) \). Answer: (d)

Teacher's Note:
1. The point of intersection of perpendicular tangents lies on the directrix.

Answer: (d) \( xy=\dfrac{105}{64} \)

Explanation:
1. Completing the square: \( y+\dfrac{35a}{16}=\dfrac{a^3}3\left(x+\dfrac3{4a}\right)^2 \), so the vertex is \( \left(-\dfrac3{4a},-\dfrac{35a}{16}\right) \).
2. The product of the coordinates is \( \dfrac3{4a}\cdot\dfrac{35a}{16}=\dfrac{105}{64} \), so the locus is \( xy=\dfrac{105}{64} \). Answer: (d)

Teacher's Note:
1. Find the vertex in terms of \( a \) and eliminate \( a \).

Answer: (d) \( y^2-4x+2=0 \)

Explanation:
1. Take \( Q=(2t^2,4t) \); the mid-point is \( (h,k)=\left(\dfrac{2t^2+1}2,2t\right) \).
2. With \( t=\dfrac k2 \): \( 2h=\dfrac{k^2}2+1 \), so \( k^2=4h-2 \), and the locus is \( y^2-4x+2=0 \). Answer: (d)

Teacher's Note:
1. Use the parametric form of the parabola.

Answer: (a) \( d^2+(2b+3c)^2=0 \)

Explanation:
1. The parabolas meet at \( (0,0) \) and \( (4a,4a) \). The line through \( (0,0) \) gives \( d=0 \); through \( (4a,4a) \) gives \( 8ab+12ac+4d=0 \), so \( 2b+3c=0 \) when \( d=0 \).
2. Squaring and adding the two conditions, \( d^2+(2b+3c)^2=0 \). Answer: (a)

Teacher's Note:
1. Substitute both intersection points into the line.

Answer: (a) \( t_2=-t_1-\dfrac2{t_1} \)

Explanation:
1. The normal at \( t_1 \) is \( y=-t_1x+2bt_1+bt_1^3 \). Putting \( (bt_2^2,2bt_2) \): \( 2t_2-2t_1=-t_1(t_2^2-t_1^2) \).
2. Dividing by \( t_2-t_1 \): \( 2=-t_1(t_2+t_1) \), so \( t_2=-t_1-\dfrac2{t_1} \). Answer: (a)

Teacher's Note:
1. Substitute the second point into the normal.

Answer: (d) \( x=3/2 \)

Explanation:
1. Rewrite as \( (y+2)^2=-4\left(x-\dfrac12\right) \), a parabola opening to the left with \( a=1 \) and vertex \( \left(\dfrac12,-2\right) \).
2. The directrix is \( x-\dfrac12=a=1 \), i.e. \( x=\dfrac32 \). Answer: (d)

Teacher's Note:
1. Complete the square to get the standard form.