JEE Mathematics Parabola MCQs Set 09

Answer: (b) \( \sqrt5 \)

Explanation:
1. The shortest segment lies along a common normal. The normal to \( y^2=4x \) at \( (t^2,2t) \) has slope \( -t \); the normal to \( y^2=2x-6 \) at a point with ordinate s has slope \( -s \), so s = t and the second point is \( \left(\frac{t^2+6}{2},t\right) \).
2. The joining line must have slope \( -t \): \( \frac{-2t}{6-t^2}=-t \), so t = 0 or \( t^2=4 \).
3. For t = 0 the points are (0, 0) and (3, 0): distance 3. For \( t=\pm2 \) the points are (4, 4) and (5, 2): distance \( \sqrt5 \). The least is \( \sqrt5 \).

Teacher's Note:
1. Nearest points of two curves lie on a common normal.
2. Compare the distances obtained for every stationary case.

Answer: (a) \( x^2-6x-8y+25=0 \)

Explanation:
1. Write the given parabola as \( (y-2)^2=12x \): vertex (0, 2), \( 4a=12 \), so a = 3 and the focus is (3, 2).
2. The required parabola has vertex (3, 2) and focus (3, 4): the axis is vertical, a = 2 and it opens upwards.
3. Its equation is \( (x-3)^2=8(y-2) \), i.e. \( x^2-6x-8y+25=0 \).

Teacher's Note:
1. Complete the square to find the focus of the given parabola.
2. The distance from vertex to focus is a.

Answer: (b) 4a

Explanation:
1. Perpendicular tangents meet on the directrix, so P = (−a, k) and AB is the chord of contact of P.
2. The chord of contact \( ky=2a(x-a) \) passes through the focus (a, 0), so AB is a focal chord.
3. The shortest focal chord is the latus rectum, of length 4a.

Teacher's Note:
1. Where do perpendicular tangents to a parabola meet?
2. Which focal chord is the shortest?

Answer: (b) \( 9y^2=4ax \)

Explanation:
1. A double ordinate is the chord \( x=at^2 \) with ends \( (at^2,\pm2at) \). Its points of trisection are \( \left(at^2,\pm\frac{2at}{3}\right) \).
2. For (x, y) = \( \left(at^2,\frac{2at}3\right) \): \( t=\frac{3y}{2a} \), so \( x=a\cdot\frac{9y^2}{4a^2} \).
3. Hence \( 9y^2=4ax \).

Teacher's Note:
1. Take the double ordinate through (at², 2at).
2. Eliminate t.

Answer: (a) \( \frac{4(\sqrt3+2)}3 \)

Explanation:
1. The line passes through P and has slope \( \sqrt3 \) (angle 60°), so points on it are \( \left(\sqrt3+\frac r2,\ \frac{\sqrt3}2r\right) \) with r = signed distance from P.
2. Substituting in the parabola: \( \frac34r^2=\sqrt3+\frac r2+2 \), i.e. \( 3r^2-2r-4(\sqrt3+2)=0 \).
3. PA . PB = |product of roots| = \( \frac{4(\sqrt3+2)}3 \).

Teacher's Note:
1. Use the parametric form of the line from P.
2. The product of the roots gives the product of the distances.

Answer: (a) always in AP

Explanation:
1. Let A and B have parameters \( t_1,t_2 \): ordinates \( 2at_1,\ 2at_2 \).
2. The tangents meet at \( C=(at_1t_2,\ a(t_1+t_2)) \), whose ordinate is \( a(t_1+t_2) \).
3. This is the arithmetic mean of \( 2at_1 \) and \( 2at_2 \), so the ordinates are in AP.

Teacher's Note:
1. Write the point of intersection of tangents at t₁ and t₂.
2. Compare the middle ordinate with the other two.

Answer: (a) isosceles

Explanation:
1. Take \( P=(at^2,2at) \). The normal \( y=-tx+2at+at^3 \) meets y = 0 at \( Q=(a(t^2+2),0) \).
2. With S = (a, 0): \( SP=a(t^2+1) \) (distance from the directrix) and \( SQ=a(t^2+2)-a=a(t^2+1) \).
3. So SP = SQ: the triangle is isosceles.

Teacher's Note:
1. Find Q from the equation of the normal.
2. SP equals the distance of P from the directrix.

Answer: (d) \( \left(1,\frac52\right) \)

Explanation:
1. The foot of the perpendicular from the focus (−1, 1) to the directrix is \( (-1,1)+\lambda(4,3) \) with \( 4(-1+4\lambda)+3(1+3\lambda)=24 \), so \( \lambda=1 \) and the foot is (3, 4).
2. The vertex is the mid-point of the focus and this foot: \( \left(\frac{-1+3}2,\frac{1+4}2\right)=\left(1,\frac52\right) \).

Teacher's Note:
1. The vertex lies midway between focus and directrix on the axis.
2. The axis is perpendicular to the directrix.

Answer: (d) x − 2y + 4 = 0

Explanation:
1. A tangent to \( y^2=4x \) is \( y=mx+\frac1m \); a tangent to \( x^2=-32y \) (a = −8) is \( y=mx+8m^2 \).
2. Equating the intercepts: \( \frac1m=8m^2 \), 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. Write the slope form of a tangent to each parabola.
2. Equal slope and equal intercept.

Answer: (c) x + 3 = 0

Explanation:
1. A tangent of slope m to the first parabola: \( y=m(x+1)+\frac1m \). A perpendicular tangent (slope \( -\frac1m \)) to the second: \( y=-\frac1m(x+2)-2m \).
2. Equating: \( m^2(x+1)+1=-(x+2)-2m^2 \), so \( (x+3)(m^2+1)=0 \).
3. Since \( m^2+1\ne0 \), the locus is x + 3 = 0.

Teacher's Note:
1. Use y = m(x − x₀) + a/m for each parabola.
2. Eliminate m between the two tangents.

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

Explanation:
1. A tangent to the parabola is \( y=mx+\frac1m \). Its distance from the centre (3, 0) must equal the radius 3: \( \frac{|3m+1/m|}{\sqrt{1+m^2}}=3 \).
2. Squaring: \( (3m^2+1)^2=9m^2(1+m^2) \), so \( 6m^2+1=9m^2 \) and \( m^2=\frac13 \).
3. Above the x-axis the intercept 1/m is positive, so \( m=\frac1{\sqrt3} \) and the tangent is \( y=\frac x{\sqrt3}+\sqrt3 \), i.e. \( \sqrt3y=x+3 \).

Teacher's Note:
1. Use the tangent y = mx + 1/m of the parabola.
2. Impose distance from the centre = radius.

Answer: (d) 90°

Explanation:
1. For \( P=(at^2,2at) \) the tangent \( ty=x+at^2 \) meets x = −a at \( K=\left(-a,\frac{a(t^2-1)}t\right) \).
2. The slope of SP is \( \frac{2t}{t^2-1} \) and the slope of SK is \( -\frac{t^2-1}{2t} \).
3. Their product is −1, so SP ⟂ SK: the angle is 90°.

Teacher's Note:
1. Find K and compare the slopes of SP and SK.
2. Product of slopes = −1 means perpendicular.

Answer: (c) \( y^2=3(x+a)^2+4ax \)

Explanation:
1. Each vertex is a point (h, k) from which two tangents are drawn at 60° to each other. Tangents \( y=mx+\frac am \) through (h, k) satisfy \( hm^2-km+a=0 \).
2. So \( (m_1-m_2)^2=\frac{k^2-4ah}{h^2} \) and \( 1+m_1m_2=\frac{h+a}h \).
3. From \( \tan60^\circ=\left|\frac{m_1-m_2}{1+m_1m_2}\right| \): \( k^2-4ah=3(h+a)^2 \). The locus is \( y^2=3(x+a)^2+4ax \).

Teacher's Note:
1. Angle between two tangents from (h, k): use tan θ = |(m₁ − m₂)/(1 + m₁m₂)|.
2. Put θ = 60°.

Answer: (d) none of these

Explanation:
1. A tangent to \( y^2=4x \): \( y=mx+\frac1m \). A tangent to \( x^2=-8y \) (a = −2): \( y=mx+2m^2 \).
2. Equating the intercepts: \( \frac1m=2m^2 \), so \( m^3=\frac12 \), \( m=2^{-1/3} \).
3. The tangent is \( y=2^{-1/3}x+2^{1/3} \), which is none of the lines listed.

Teacher's Note:
1. Same method: slope form for both parabolas.
2. Check whether m = 2^(−1/3) matches any option.

Answer: (c) 32

Explanation:
1. Here \( 4a=32 \), so a = 8.
2. A focal chord has length \( a\left(t+\frac1t\right)^2 \), which is at least 4a, with equality for the latus rectum.
3. So PQ is never less than 32.

Teacher's Note:
1. Length of a focal chord with end parameters t and −1/t.
2. (t + 1/t)² ≥ 4.

Answer: (a) \( (-2+2\sqrt2,\ 2) \)

Explanation:
1. The chord joining (2, 2) and (8, −4) is x + y = 4. The region lies inside the parabola and on the side of the chord containing the vertex: \( y^2<2x \) and \( x+y<4 \).
2. Inside the parabola: \( (\alpha-2)^2<2\alpha^2 \), i.e. \( \alpha^2+4\alpha-4>0 \), so \( \alpha>-2+2\sqrt2 \) or \( \alpha<-2-2\sqrt2 \).
3. Same side as the vertex: \( \alpha^2+\alpha-2<4 \), i.e. \( -3<\alpha<2 \). Combining: \( \alpha\in(-2+2\sqrt2,\ 2) \).

Teacher's Note:
1. Write both inequalities for the region.
2. Intersect the two solution sets.

Answer: (d) \( T^2\ge8 \)

Explanation:
1. For a normal at parameter t meeting the parabola again at T: \( T=-t-\frac2t \), i.e. \( t^2+Tt+2=0 \).
2. For real t the discriminant must be non-negative: \( T^2-8\ge0 \).
3. So \( T^2\ge8 \).

Teacher's Note:
1. Recall T = −t − 2/t.
2. Real t needs a non-negative discriminant.

Answer: (c) \( \frac\pi2 \)

Explanation:
1. For a focal chord the parameters satisfy \( t_1t_2=-1 \).
2. The tangents have slopes \( \frac1{t_1} \) and \( \frac1{t_2} \), with product \( \frac1{t_1t_2}=-1 \).
3. So the tangents are perpendicular and \( \angle ACB=\frac\pi2 \).

Teacher's Note:
1. What is t₁t₂ for a focal chord?
2. The slope of the tangent at t is 1/t.

Answer: (a) \( \frac1a \)

Explanation:
1. With parameters t and −1/t: \( SP=a(t^2+1) \) and \( SQ=a\left(\frac1{t^2}+1\right)=\frac{a(1+t^2)}{t^2} \).
2. Then \( \frac1{SP}+\frac1{SQ}=\frac1{a(t^2+1)}+\frac{t^2}{a(t^2+1)}=\frac1a \).

Teacher's Note:
1. Distance from the focus = a(t² + 1).
2. Add the reciprocals.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Four concyclic points on the parabola have parameters with \( t_1+t_2+t_3+t_4=0 \) (the sum of the ordinates is zero), so R is true.
2. The normal at A meets the parabola again at B: \( t_1+t_2=-\frac2{t_1} \); likewise \( t_3+t_4=-\frac2{t_3} \). Adding gives \( \frac1{t_1}+\frac1{t_3}=0 \), so \( t_3=-t_1 \).
3. The tangents at A and C meet at \( (at_1t_3,\ a(t_1+t_3))=(-at_1^2,\ 0) \), a point on the axis, so A is true and R is its explanation.

Teacher's Note:
1. Concyclic points on a parabola satisfy Σt = 0.
2. Use the normal-chord relation t₂ = −t₁ − 2/t₁.