JEE Mathematics Parabola MCQs Set 12

Answer: (d) \( (4,0) \)

Explanation:
1. The axis is perpendicular to the directrix, so the axis is the x-axis.
2. The vertex is midway between focus and directrix, and the directrix x = 0 is 2 units from the vertex (2, 0).
3. So the focus is 2 units on the other side of the vertex: (4, 0).

Teacher's Note:
1. The vertex bisects the perpendicular from focus to directrix.
2. Focus = 2 × vertex − foot of perpendicular on the directrix = (4, 0) − (0, 0).

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

Explanation:
1. Rewrite the given parabola as \( (y-2)^2=12x \): vertex (0, 2), 4a = 12, so a = 3 and the focus is (3, 2).
2. The new parabola has vertex (3, 2) and focus (3, 4): the axis is vertical and a = 2, opening upwards.
3. Equation: \( (x-3)^2=8(y-2) \), i.e. \( x^2-6x-8y+25=0 \).

Teacher's Note:
1. Complete the square in y to read off the vertex and a.
2. The new axis is the line x = 3, so the squared term is in x.

Answer: (d) \( (1,2\sqrt3),(1,-2\sqrt3) \)

Explanation:
1. Here 4a = 12, so a = 3. The focal distance of a point (x, y) is x + a = x + 3.
2. x + 3 = 4 gives x = 1.
3. Then \( y^2=12 \), so \( y=\pm2\sqrt3 \).

Teacher's Note:
1. Focal distance of any point on y² = 4ax equals its distance from the directrix, x + a.
2. Check: the point (1, 2√3) lies on y² = 12x since 12 = 12.

Answer: (d) \( (-5,5) \)

Explanation:
1. Outside the parabola (the side not containing the focus, y < x² + x + 1): \( \alpha<\alpha^2+\alpha+1 \), i.e. \( \alpha^2+1>0 \), true for every real \( \alpha \).
2. Inside the circle: \( 2\alpha^2<50 \), so \( |\alpha|<5 \).
3. Hence \( \alpha\in(-5,5) \).

Teacher's Note:
1. Substitute the point into the expression y − x² − x − 1: a negative value means outside this upward-opening parabola.
2. The first condition is always true, so only the circle restricts alpha.

Answer: (d) a parabola

Explanation:
1. Take the base from (−a, 0) to (a, 0) and the vertex (x, y). The tangents of the base angles are \( \frac{y}{x+a} \) and \( \frac{y}{a-x} \).
2. Their sum is \( \frac{2ay}{a^2-x^2}=k \).
3. So \( x^2=a^2-\frac{2a}{k}y \), which is a parabola.

Teacher's Note:
1. Place the base on the x-axis and write each tangent as height over horizontal distance.
2. The equation has x² but only y to the first power, so it is a parabola.

Answer: (b) \( \frac{2u^2}{g}\cos^2\alpha \)

Explanation:
1. The distance from the focus to the directrix is \( \frac{u^2}{2g}+\frac{u^2}{2g}\cos2\alpha=\frac{u^2}{2g}(1+\cos2\alpha)=\frac{u^2}{g}\cos^2\alpha \).
2. The latus rectum is twice the focus-directrix distance (4a = 2 × 2a).
3. So the length is \( \frac{2u^2}{g}\cos^2\alpha \).

Teacher's Note:
1. Latus rectum = 2 × (distance from focus to directrix).
2. Use 1 + cos 2α = 2cos²α.

Answer: (a) 12

Explanation:
1. Write \( u=\frac{\sqrt3x-y}{2} \) and \( v=\frac{x+\sqrt3y}{2} \); these are perpendicular distances from two perpendicular lines through the origin.
2. The equation becomes \( 4u^2=96v \), i.e. \( u^2=24v \).
3. So 4a = 24, a = 6, and the focus-directrix distance is 2a = 12.

Teacher's Note:
1. Divide the linear expression by its norm (here 2) so that it measures real distance.
2. The two lines are perpendicular, so u and v act as rotated coordinate axes.

Answer: (b) \( \pi/2 \)

Explanation:
1. A double ordinate of length 8a has ends \( (4a,\ 4a) \) and \( (4a,-4a) \) (since y = ±4a gives x = 4a).
2. Each end makes an angle 45° with the axis, so the angle subtended at the vertex is 90°.

Teacher's Note:
1. Find the ends by putting y = 4a in y² = 4ax.
2. Slopes of the two lines from the vertex are +1 and −1, whose product is −1.

Answer: (c) \( \frac{24}{5} \)

Explanation:
1. For a focal chord, the semi-latus rectum is the harmonic mean of the focal segments: \( \frac1{SP}+\frac1{SQ}=\frac1a \).
2. So \( \frac1a=\frac13+\frac12=\frac56 \) and \( a=\frac65 \).
3. Latus rectum = 4a = \( \frac{24}5 \).

Teacher's Note:
1. Segments of a focal chord satisfy 1/SP + 1/SQ = 1/a, where 4a is the latus rectum.
2. Here a is the quantity in y² = 4ax.

Answer: (a) \( \frac{16}{3} \)

Explanation:
1. The parabola has vertex (1, 0) and a = 1, so its focus is (2, 0).
2. The line passes through (2, 0), so the chord is a focal chord. Its slope is \( \sqrt3 \), so it makes 60° with the axis.
3. A focal chord inclined at angle \( \theta \) has length \( \frac{4a}{\sin^2\theta}=\frac{4}{3/4}=\frac{16}{3} \).

Teacher's Note:
1. Check whether the line passes through the focus before doing a long calculation.
2. Focal chord length = 4a / sin²θ.

Answer: (d) \( -\frac{14}{3} \)

Explanation:
1. Put \( X=x-\tfrac13 \): the parabola is \( y^2=4aX \) and the line becomes \( y=2X-\tfrac73 \).
2. Tangency for y = mX + c requires c = a/m: \( -\tfrac73=\tfrac a2 \).
3. So \( a=-\tfrac{14}{3} \).

Teacher's Note:
1. Shift the origin to the vertex before applying the tangent condition.
2. The tangent condition for y² = 4aX is c = a/m.

Answer: (d) \( \pi/2 \)

Explanation:
1. The point (−a, 2a) lies on the directrix x = −a.
2. Tangents drawn from any point of the directrix are perpendicular (the directrix is the director circle of a parabola).
3. So the angle is \( \pi/2 \).

Teacher's Note:
1. Check the x-coordinate against the directrix first.
2. Algebra check: the tangent y = mx + a/m through (−a, 2a) gives m² + 2m − 1 = 0 with product of slopes −1.

Answer: (a) \( \left(\frac52,\frac72\right) \)

Explanation:
1. AB is the chord of contact of the required point (h, k). Its equation is \( ky=2(x+h) \), i.e. \( 2x-ky+2h=0 \).
2. Compare with \( 4x-7y+10=0 \): \( \frac24=\frac{-k}{-7}=\frac{2h}{10} \).
3. So \( k=\frac72,\ h=\frac52 \).

Teacher's Note:
1. The chord of contact from (h, k) to y² = 4x is yk = 2(x + h).
2. Equate ratios of coefficients.

Answer: (d) \( y^2=2(x-1) \)

Explanation:
1. Let the mid-point be (h, k). Using T = S₁: \( ky-2(x+h)=k^2-4h \).
2. The chord passes through the focus (1, 0): \( -2(1+h)=k^2-4h \), so \( k^2=2h-2 \).
3. Locus: \( y^2=2(x-1) \).

Teacher's Note:
1. Use the equation of a chord with a given mid-point, T = S₁.
2. Force it through the focus.

Answer: (b) (1, 2)

Explanation:
1. Equal angles with the axes means the normal has slope ±1.
2. The normal at \( (t^2,2t) \) has slope \( -t \), so \( t=\pm1 \).
3. The points are (1, 2) and (1, −2); (1, 2) is listed.

Teacher's Note:
1. Slope of the normal at the point with parameter t on y² = 4x is −t.
2. Only (1, 2) is among the options.

Answer: (a) 24

Explanation:
1. Here 4a = −8, so a = −2. For \( y^2=4ax \) the normal with slope m is \( y=mx-2am-am^3 \).
2. The line \( y=-2x-\lambda \) has slope m = −2, so its intercept is \( -2am-am^3=-2(-2)(-2)-(-2)(-8)=-8-16=-24 \).
3. Hence \( -\lambda=-24 \) and \( \lambda=24 \).

Teacher's Note:
1. Use the normal in slope form with a = −2 (the parabola opens to the left).
2. Check: the normal touches the curve at (am², −2am) = (−8, −8), which lies on y² = −8x and on the line 2x + y + 24 = 0.

Answer: (c) \( 5x+y-135a=0 \)

Explanation:
1. The points correspond to \( t=-2 \) and \( t=-3 \) (since \( (at^2,2at) \)).
2. Three normals through one point have parameters summing to zero, so the third is \( t_3=5 \).
3. The normal at t: \( y+tx=2at+at^3 \). With t = 5: \( y+5x=10a+125a=135a \), i.e. \( 5x+y-135a=0 \).

Teacher's Note:
1. For concurrent normals on a parabola, t₁ + t₂ + t₃ = 0.
2. Then write the normal at the third parameter.

Answer: (c) 2

Explanation:
1. Here 4a = 1, so \( a=\tfrac14 \). The normal \( y=mx-2am-am^3 \) through (c, 0) gives \( m\,(c-2a-am^2)=0 \).
2. One root is m = 0; two more real distinct roots need \( m^2=\frac{c-2a}{a}>0 \).
3. So \( c>2a=\tfrac12 \). Only c = 2 qualifies.

Teacher's Note:
1. From a point on the axis, one normal is always the axis itself (m = 0).
2. Three real normals need c > 2a.

Answer: (c) \( y+4=0 \)

Explanation:
1. The feet of the perpendiculars from the focus on the tangents lie on the tangent at the vertex.
2. Write \( y+4=(x-1)^2 \): the vertex is (1, −4).
3. The tangent at the vertex is horizontal: \( y=-4 \), i.e. \( y+4=0 \).

Teacher's Note:
1. The tangent at the vertex is the locus of feet of perpendiculars from the focus.
2. Complete the square to find the vertex.

Answer: (b) \( (x-a)^2=\tfrac12(2y-2b) \)

Explanation:
1. With vertex (a, b) and the axis along +y, the equation is \( (x-a)^2=4A(y-b) \) where 4A is the latus rectum.
2. Latus rectum = 1, so \( (x-a)^2=1\cdot(y-b)= frac12(2y-2b) \).

Teacher's Note:
1. The coefficient of (y − b) equals the length of the latus rectum.
2. Match with the second option.

Answer: (d) \( \arctan\sqrt2 \)

Explanation:
1. For the parabola \( y^2=4ax \), the normal at the point \( t \) meets the curve again at \( t_2=-t-rac2t \).
2. Right angle at the vertex: \( t\,t_2=-4 \), so \( -t^2-2=-4 \) and \( t^2=2 \).
3. The normal has slope \( -t \), so \( \tan\theta=\sqrt2 \) for the acute angle.

Teacher's Note:
1. Perpendicular chords from the vertex: product of parameters is −4.
2. The slope of the normal at t is −t.

Answer: (b) \( (a,0) \)

Explanation:
1. Let the feet be \( (at_1^2,2at_1) \) and \( (at_2^2,2at_2) \). The normals have slopes \( -t_1,-t_2 \), so perpendicular means \( t_1t_2=-1 \).
2. The chord joining the feet: \( 2x-(t_1+t_2)y+2at_1t_2=0 \).
3. With \( t_1t_2=-1 \) it becomes \( 2x-(t_1+t_2)y-2a=0 \), which passes through \( (a,0) \) for all t₁ + t₂.

Teacher's Note:
1. Write the chord through two points of the parabola in terms of t₁ and t₂.
2. The fixed point makes the coefficient of the varying term vanish: y = 0.

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

Explanation:
1. Let the focus be (h, k). The feet of the perpendiculars from the focus on the two tangents (the axes) are (h, 0) and (0, k), and both lie on the tangent at the vertex.
2. That tangent is \( \frac xh+\frac yk=1 \).
3. Its distance from the focus is a: \( \frac{1}{\sqrt{1/h^2+1/k^2}}=a \), so \( h^2k^2=a^2(h^2+k^2) \).

Teacher's Note:
1. Feet of perpendiculars from the focus on tangents lie on the vertex tangent.
2. Distance from focus to the tangent at the vertex equals a.

Answer: (d) 4a

Explanation:
1. With P and Q having parameters \( t_1,t_2 \), perpendicularity at the vertex gives \( t_1t_2=-4 \).
2. The chord PQ: \( 2x-(t_1+t_2)y+2at_1t_2=0 \). Put y = 0: \( x=-at_1t_2=4a \).
3. So OR = 4a.

Teacher's Note:
1. The chord through two points of the parabola is 2x − (t₁ + t₂)y + 2at₁t₂ = 0.
2. The x-intercept does not depend on t₁ + t₂.

Answer: (c) \( y^2=4a(x-8a) \)

Explanation:
1. Let the chords go to \( (at_1^2,2at_1) \) and \( (at_2^2,2at_2) \) with \( t_1t_2=-4 \).
2. The fourth vertex is the sum: \( x=a(t_1^2+t_2^2)=a\{(t_1+t_2)^2+8\} \), \( y=2a(t_1+t_2) \).
3. Then \( y^2=4a^2(t_1+t_2)^2=4a(x-8a) \).

Teacher's Note:
1. In a rectangle OPRQ the fourth vertex R is P + Q (as vectors from O).
2. Use (t₁ + t₂)² = t₁² + t₂² + 2t₁t₂ = t₁² + t₂² − 8.

Answer: (d) \( \frac12 \)

Explanation:
1. A tangent to \( y^2=4x \) is \( y=mx+\frac1m \). A tangent to \( x^2=-32y \) is \( y=mx+8m^2 \) (here a = −8: \( y=mx-am^2 \) with \( 4a=-32 \)).
2. Common tangent: \( \frac1m=8m^2 \), so \( m^3=\frac18 \).
3. Hence \( m=\frac12 \).

Teacher's Note:
1. For x² = 4ay the tangent is y = mx − am².
2. Equate the intercepts of the two tangent forms.

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

Explanation:
1. Let Q = (h, k) with \( h^2=8k \). P divides OQ in ratio 1 : 3, so \( P=(x,y)=\left(\frac h4,\frac k4\right) \).
2. Then \( h=4x,\ k=4y \) and \( 16x^2=32y \).
3. The locus is \( x^2=2y \).

Teacher's Note:
1. Use the section formula for P between O (origin) and Q.
2. Substitute Q = (4x, 4y) in the parabola.

Answer: (d) \( x^2+y^2-4x+8y+12=0 \)

Explanation:
1. C = (0, −6). The nearest point of the parabola to C lies where the normal passes through C. The normal to \( y^2=8x \) (a = 2) is \( y=-mx+4m+2m^3 \); through (0, −6): \( 2m^3+4m+6=0 \), i.e. \( m^3+2m+3=0 \), so m = −1.
2. P = (2m², 4m) = (2, −4). Radius² = PC² = 4 + 4 = 8.
3. Circle: \( (x-2)^2+(y+4)^2=8 \), i.e. \( x^2+y^2-4x+8y+12=0 \).

Teacher's Note:
1. The shortest distance from a point to a curve is along a normal.
2. Only m = −1 is a real root of m³ + 2m + 3 = 0.