JEE Mathematics Parabola MCQs Set 18

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

Explanation:
1. For a point \( (x, y) \) on the parabola, distance from focus = distance from directrix.
2. \( (x+3)^2 + y^2 = (x+5)^2 \).
3. \( x^2 + 6x + 9 + y^2 = x^2 + 10x + 25 \), so \( y^2 = 4x + 16 = 4(x+4) \).

Teacher's Note:
1. Use the focus-directrix definition directly.
2. Squaring removes the modulus on the directrix distance.

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

Explanation:
1. The axis is perpendicular to the directrix, so it is the x-axis, meeting the directrix at \( (0, 0) \).
2. The vertex is the midpoint of the foot of the directrix and the focus.
3. So the focus is \( (4, 0) \).

Teacher's Note:
1. The vertex is halfway between focus and directrix.
2. The focus lies on the axis of the parabola.

Answer: (d) \( \frac{2}{5} \)

Explanation:
1. Rewrite as \( (x-2)^2 + (y-3)^2 = \left(\frac{3x-4y+7}{5}\right)^2 \): focus \( (2,3) \), directrix \( 3x - 4y + 7 = 0 \).
2. Distance from focus to directrix \( = \frac{|6 - 12 + 7|}{5} = \frac{1}{5} \).
3. Latus rectum \( = 2 \times \frac{1}{5} = \frac{2}{5} \).

Teacher's Note:
1. The latus rectum equals twice the focus-directrix distance.
2. Spot the focus-directrix form by dividing the line by \( \sqrt{3^2 + 4^2} = 5 \).

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

Explanation:
1. \( y^2 - 4y = 12x - 4 \) gives \( (y-2)^2 = 12x \): vertex \( (0, 2) \), \( a = 3 \), focus \( (3, 2) \).
2. New parabola: vertex \( (3, 2) \), focus \( (3, 4) \), so it opens upward with \( a = 2 \).
3. \( (x-3)^2 = 8(y-2) \), i.e. \( x^2 - 6x - 8y + 25 = 0 \).

Teacher's Note:
1. Complete the square to find the first focus.
2. The distance from vertex to focus gives \( a \).

Answer: (b) \( x^2 - 2 = -2\cos t,\ y = 4\cos^2 \frac{t}{2} \)

Explanation:
1. In (b), \( x^2 = 2 - 2\cos t \) and \( y = 2(1 + \cos t) = 2 + 2\cos t \).
2. Adding: \( x^2 + y = 4 \), i.e. \( y = 4 - x^2 \), a parabola.
3. (a) is an ellipse; (c) gives \( y - x = 1 \), a line; (d) gives \( y^2 = 1 + \sin t = 2 - x^2 \), a circle.

Teacher's Note:
1. Eliminate the parameter to identify the curve.
2. A second-degree equation in only one variable is a parabola.

Answer: (b) \( (1, 2\sqrt{3}),\ (1, -2\sqrt{3}) \)

Explanation:
1. Here \( a = 3 \). For \( P(3t^2, 6t) \), focal distance \( = x + a = 3t^2 + 3 \).
2. \( 3t^2 + 3 = 4 \) gives \( t^2 = \frac{1}{3} \).
3. So \( x = 1,\ y = \pm 2\sqrt{3} \).

Teacher's Note:
1. Focal distance of a point is its x-coordinate plus \( a \).
2. Check the answer lies on \( y^2 = 12x \): \( 12 = 12 \).

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

Explanation:
1. Outside the parabola (below it) needs \( \alpha < \alpha^2 + \alpha + 1 \), i.e. \( \alpha^2 + 1 > 0 \), true for all real \( \alpha \).
2. Inside the circle needs \( 2\alpha^2 < 50 \), so \( \alpha^2 < 25 \).
3. Hence \( \alpha \in (-5, 5) \).

Teacher's Note:
1. Substitute the point into \( y - (x^2 + x + 1) \) to test its side.
2. Combine both conditions by intersection.

Answer: (b) a parabola

Explanation:
1. Take the base from \( A(-a, 0) \) to \( B(a, 0) \) and vertex \( P(h, k) \).
2. \( \tan\alpha + \tan\beta = \frac{k}{h+a} + \frac{k}{a-h} = \frac{2ak}{a^2 - h^2} = \lambda \).
3. So \( 2ak = \lambda(a^2 - h^2) \), i.e. \( h^2 = a^2 - \frac{2ak}{\lambda} \): a parabola.

Teacher's Note:
1. Write the tangents of the base angles using the coordinates of the vertex.
2. An equation linear in \( y \) and quadratic in \( x \) is a parabola.

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

Explanation:
1. Distance from focus to directrix \( = \frac{u^2}{2g} + \frac{u^2}{2g}\cos 2\alpha = \frac{u^2}{2g}(1 + \cos 2\alpha) = \frac{u^2}{g}\cos^2\alpha \).
2. Latus rectum is twice this distance.
3. So it equals \( \frac{2u^2}{g}\cos^2\alpha \).

Teacher's Note:
1. The directrix is horizontal, so only the y-coordinates matter.
2. Use \( 1 + \cos 2\alpha = 2\cos^2\alpha \).

Answer: (d) 12

Explanation:
1. Divide by 4: \( \left(\frac{\sqrt{3}x - y}{2}\right)^2 = 24\cdot\frac{x + \sqrt{3}y}{2} \).
2. The two brackets are distances from the perpendicular lines \( \sqrt{3}x - y = 0 \) (axis) and \( x + \sqrt{3}y = 0 \) (tangent at vertex): \( Y^2 = 24X \).
3. So \( 4a = 24 \), \( a = 6 \), and the focus-directrix distance is \( 2a = 12 \).

Teacher's Note:
1. Rotate axes mentally: the brackets are perpendicular distances.
2. Compare with \( Y^2 = 4aX \).

Answer: (c) \( xy - 2 = 0 \)

Explanation:
1. \( (1, 2) \) is an end of the latus rectum, so the focal chord is the latus rectum and the other end is \( (1, -2) \).
2. (a): \( -2 + 2 = 0 \) holds; (b): \( -2 + 2 = 0 \) holds; (d): \( 1 - 2 + 2 - 1 = 0 \) holds.
3. (c): \( -2 - 2 = -4 \ne 0 \), so the point does not lie on it.

Teacher's Note:
1. The chord through the focus and \( (1, 2) \) is vertical.
2. Test the candidate point in each curve.

Answer: (b) \( \frac{\pi}{2} \)

Explanation:
1. Half the double ordinate is \( 4a \), so the end points are \( (x, \pm 4a) \).
2. \( 16a^2 = 4ax \) gives \( x = 4a \): points \( (4a, 4a) \) and \( (4a, -4a) \).
3. Slopes from the vertex are \( 1 \) and \( -1 \), product \( -1 \), so the angle is \( \frac{\pi}{2} \).

Teacher's Note:
1. Find the abscissa of the double ordinate first.
2. Perpendicular lines have slopes whose product is \( -1 \).

Answer: (b) GP

Explanation:
1. At \( (x_1, y_1) \): subtangent \( = 2x_1 \), ordinate \( = y_1 \), subnormal \( = 2a \).
2. \( y_1^2 = 4ax_1 = (2x_1)(2a) \), so the ordinate is the geometric mean.
3. Hence they are in GP.

Teacher's Note:
1. Recall the standard lengths for \( y^2 = 4ax \).
2. Square the middle term and compare with the product of the others.

Answer: (a) \( y = -mx + 2am + am^3 \)

Explanation:
1. \( \frac{dy}{dx} = \frac{2a}{y} = \frac{1}{m} \) at the point, so the normal slope is \( -m \).
2. \( y - 2am = -m(x - am^2) \).
3. So \( y = -mx + 2am + am^3 \).

Teacher's Note:
1. Normal slope is the negative reciprocal of tangent slope.
2. Memorise \( y = mx - 2am - am^3 \) for the point \( (am^2, -2am) \) too.

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

Explanation:
1. Equal angles with the axes means slope \( \pm 1 \).
2. \( \frac{dy}{dx} = \frac{2}{y} \); the normal slope is \( -\frac{y}{2} = \pm 1 \), so \( y = \pm 2 \).
3. \( y = 2 \) gives \( x = 1 \): the point \( (1, 2) \) lies on the parabola and the normal has slope \( -1 \).

Teacher's Note:
1. Slope \( \pm 1 \) is the key condition.
2. Only \( (1, 2) \) among the options has \( |y| = 2 \).

Answer: (c) 24

Explanation:
1. For \( y^2 = -8x \), \( 2y\,y' = -8 \), so the tangent slope is \( -\frac{4}{y_0} \) and normal slope \( \frac{y_0}{4} \).
2. The line has slope \( -2 \): \( y_0 = -8 \), \( x_0 = -8 \).
3. The line passes through \( (-8, -8) \): \( -16 - 8 + \lambda = 0 \), so \( \lambda = 24 \).

Teacher's Note:
1. Match the slope of the normal to find the point of contact.
2. Substitute the point into the line to get \( \lambda \).

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

Explanation:
1. The points are \( (am^2, -2am) \) with \( m_1 = 2 \) and \( m_2 = 3 \).
2. The slopes of three concurrent normals satisfy \( m_1 + m_2 + m_3 = 0 \), so \( m_3 = -5 \).
3. Normal: \( y = m_3x - 2am_3 - am_3^3 = -5x + 10a + 125a \), i.e. \( 5x + y - 135a = 0 \).

Teacher's Note:
1. For \( y^2 = 4ax \), the parameters of three concurrent normals add up to zero.
2. Use the normal in slope form.

Answer: (d) 2

Explanation:
1. The normal at \( (h^2, h) \) is \( y - h = -2h(x - h^2) \).
2. Passing through \( (c, 0) \): \( -h = -2h(c - h^2) \), so \( h = 0 \) or \( h^2 = c - \frac{1}{2} \).
3. Three normals need two more distinct real \( h \), i.e. \( c > \frac{1}{2} \). Only 2 qualifies.

Teacher's Note:
1. One normal is always the axis \( (h = 0) \).
2. Two further normals need a positive value of \( h^2 \).

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

Explanation:
1. For a focal chord, \( \frac{1}{SP} + \frac{1}{SQ} = \frac{1}{a} \).
2. \( \frac{1}{3} + \frac{1}{2} = \frac{5}{6} = \frac{1}{a} \), so \( a = \frac{6}{5} \).
3. Latus rectum \( = 4a = \frac{24}{5} \).

Teacher's Note:
1. The harmonic-mean property of focal chords is worth memorising.
2. The semi-latus rectum is the harmonic mean of the focal segments.

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

Explanation:
1. The line is \( y = \sqrt{3}(x - 2) \), passing through the focus \( (2, 0) \) of \( y^2 = 4(x-1) \); it is a focal chord of slope \( \sqrt{3} \), so \( \alpha = 60^\circ \).
2. Focal chord length \( = 4a\,\text{cosec}^2\alpha = 4 \times \frac{4}{3} \).
3. So the length is \( \frac{16}{3} \).

Teacher's Note:
1. Check whether the line passes through the focus first.
2. A focal chord inclined at \( \alpha \) has length \( 4a\,\text{cosec}^2\alpha \).