JEE Mathematics Parabola MCQs Set F

Question. Latus rectum of the parabola whose focus is \( (3, 4) \) and whose tangent at vertex has the equation \( x + y = 7 + 5\sqrt{2} \) is
(a) 5
(b) 10
(c) 20
(d) 15
Answer: (c) 20
Solution:
\( a = \perp^r \) distance from \( (3, 4) \) to the tangent at vertex
\( = \left| \frac{3 + 4 - 7 - 5\sqrt{2}}{\sqrt{2}} \right| \)
\( a = 5 \)
\( LR = 4a = 20 \)

Question. Directrix of a parabola is \( x + y = 2 \). If its focus is origin, then latus rectum of the parabola is equal to
(a) \( \sqrt{2} \) units
(b) 2 units
(c) \( 2\sqrt{2} \) units
(d) 4 units
Answer: (c) \( 2\sqrt{2} \) units
Solution:
Directrix : \( x + y - 2 = 0 \)
Focus to directrix distance = \( 2a \)
\( 2a = \left| \frac{0 + 0 - 2}{\sqrt{2}} \right| \)
\( 2a = \sqrt{2} \)
\( LR = 4a = 2\sqrt{2} \)

Question. Which one of the following equations represents parametrically, parabolic profile ?
(a) \( x = 3 \cos t ; y = 4 \sin t \)
(b) \( x^2 - 2 = - \cos t ; y = 4 \cos^2 \frac{t}{2} \)
(c) \( \sqrt{x} = \tan t ; \sqrt{y} = \sec t \)
(d) \( x = \sqrt{1 - \sin t} ; y = \sin \frac{t}{2} + \cos \frac{t}{2} \)
Answer: (b) \( x^2 - 2 = - \cos t ; y = 4 \cos^2 \frac{t}{2} \)
Solution:
(a) \( \frac{x^2}{9} + \frac{y^2}{16} = 1 \) Ellipse
(b) \( x^2 - 2 = - \left( 2\cos^2 \frac{t}{2} - 1 \right) \)
\( x^2 - 2 = - \left( \frac{y}{2} - 1 \right) \)
\( x^2 - 2 = - \frac{y}{2} + 1 \)
\( x^2 = - \frac{y}{2} + 3 \)
\( x^2 = - \frac{1}{2}(y - 6) \) Parabola

Question. Let C be a circle and L a line on the same plane such that C and L do not intersect. Let P be a moving point such that the circle drawn with centre at P to touch L also touches C. Then the locus of P is
(a) a straight line parallel to L not intersecting C
(b) a circle concentric with C
(c) a parabola whose focus is centre of C and whose directrix is L.
(d) a parabola whose focus is the centre of C and whose directrix is a straight line parallel to L.
Answer: (d) a parabola whose focus is the centre of C and whose directrix is a straight line parallel to L.
Solution:
Locus of P will be parabola.

Question. If \( (t^2, 2t) \) is one end of a focal chord of the parabola \( y^2 = 4x \) then the length of the focal chord will be
(a) \( \left(t + \frac{1}{t}\right)^2 \)
(b) \( \left(t + \frac{1}{t}\right) \sqrt{t^2 + \frac{1}{t^2}} \)
(c) \( \left(t - \frac{1}{t}\right) \sqrt{t^2 + \frac{1}{t^2}} \)
(d) None of the options
Answer: (a) \( \left(t + \frac{1}{t}\right)^2 \)
Solution:
\( y^2 = 4x \), \( a = 1 \)
\( P(t^2, 2t) \), \( t_1 t_2 = -1 \) For focal chord
\( t_2 = - \frac{1}{t} \)
\( Q \left( \frac{1}{t^2}, -\frac{2}{t} \right) \)
\( PQ = \sqrt{ \left(t^2 - \frac{1}{t^2}\right)^2 + \left(2t + \frac{2}{t}\right)^2 } \)
\( = \left(t + \frac{1}{t}\right) \sqrt{ \left(t - \frac{1}{t}\right)^2 + 4 } \)
\( = \left(t + \frac{1}{t}\right) \sqrt{ \left(t + \frac{1}{t}\right)^2 } \)
\( = \left(t + \frac{1}{t}\right)^2 \)

Question. From the focus of the parabola \( y^2 = 8x \) as centre, a circle is described so that a common chord of the curves is equidistant from the vertex and focus of the parabola. The equation of the circle is
(a) \( (x - 2)^2 + y^2 = 3 \)
(b) \( (x - 2)^2 + y^2 = 9 \)
(c) \( (x + 2)^2 + y^2 = 9 \)
(d) None of the options
Answer: (b) \( (x - 2)^2 + y^2 = 9 \)
Solution:
Intersection point of \( x = 1 \) with \( y^2 = 8x \)
\( P(1, 2\sqrt{2}) \)
\( r^2 = SP^2 \)
\( = (1 - 2)^2 + (2\sqrt{2})^2 \)
\( = 1 + 8 = 9 \)
equation of circle as centre \( (2, 0) \) ; \( r = 3 \)
\( (x - 2)^2 + y^2 = 9 \)

Question. The point of intersection of the curves whose parametric equations are \( x = t^2 + 1, y = 2t \) and \( x = 2s, y = 2/s \) is given by
(a) \( (1, -3) \)
(b) \( (2, 2) \)
(c) \( (-2, 4) \)
(d) \( (1, 2) \)
Answer: (b) \( (2, 2) \)
Solution:
\( x = t^2 + 1 ; y = 2t \)
\( \implies \) \( t = \frac{y}{2} \)
\( x = \frac{y^2}{4} + 1 \dots (i) \)
\( x = 2s ; y = \frac{2}{s} \)
\( \implies \) \( s = \frac{2}{y} \)
\( x = \frac{4}{y} \)
\( \frac{4}{y} = \frac{y^2}{4} + 1 \)
\( y^3 + 4y - 16 = 0 \)
\( \implies \) \( y = 2, x = 2 \) POI
Aliter
Assume a point on hyperbola \( \left(2t, \frac{2}{t}\right) \)
Put in parabola
\( 2t = \frac{1}{t^2} + 1 \)
\( 2t^3 - t^2 - 1 = 0 \)
\( t = 1 \) will satisfy point \( (2, 2) \)

Question. If M is the foot of the perpendicular from a point P of a parabola \( y^2 = 4ax \) to its directrix and SPM is an equilateral triangle, where S is the focus, then SP is equal to
(a) \( a \)
(b) \( 2a \)
(c) \( 3a \)
(d) \( 4a \)
Answer: (d) \( 4a \)
Solution:
\( PM = SM \)
\( PM^2 = SM^2 \)
\( (a + at^2)^2 = 4a^2 + 4a^2t^2 \)
\( a^2 + a^2t^4 + 2a^2t^2 = 4a^2 + 4a^2t^2 \)
\( 1 + t^4 + 2t^2 = 4 + 4t^2 \)
\( t^4 - 2t^2 - 3 = 0 \)
\( t^2 = 3, t^2 = - 1 \)
\( SP = a + at^2 \)
\( = a + 3a = 4a \)

Question. Through the vertex 'O' of the parabola \( y^2 = 4ax \), variable chords OP and OQ are drawn at right angles. If the variable chord PQ intersects the axis of x at R, then distance OR
(a) varies with different positions of P and Q
(b) equals the semi latus rectum of the parabola
(c) equals latus rectum of the parabola
(d) equals double the latus rectum of the parabola
Answer: (c) equals latus rectum of the parabola
Solution:
\( m_{OP} = \frac{2}{t_1} \)
\( m_{OQ} = \frac{2}{t_2} \)
\( \frac{2}{t_1} \times \frac{2}{t_2} = - 1 \)
\( \implies \) \( t_1 t_2 = - 4 \dots(i) \)
Equation of PQ
\( 2x - (t_1 + t_2)y + 2at_1 t_2 = 0 \)
put \( y = 0 \)
\( x = - at_1 t_2 \)
\( x = - a(-4) \)
\( x = 4a \)
\( R(4a, 0) \)

Question. The triangle PQR of area 'A' is inscribed in the parabola \( y^2 = 4ax \) such that the vertex P lies at the vertex of the parabola and the base OR is a focal chord. The modulus of the difference of the ordinates of the points Q and R is
(a) \( \frac{A}{2a} \)
(b) \( \frac{A}{a} \)
(c) \( \frac{2A}{a} \)
(d) \( \frac{4A}{a} \)
Answer: (c) \( \frac{2A}{a} \)
Solution:
\( A = \frac{1}{2} \begin{vmatrix} 0 & 0 & 1 \\ at_1^2 & 2at_1 & 1 \\ at_2^2 & 2at_2 & 1 \end{vmatrix} \)
\( = \frac{1}{2} \left| 2a^2t_1^2t_2 - 2a^2t_1t_2^2 \right| \)
\( = \frac{1}{2} \left| at_1t_2 [2at_1 - 2at_2] \right| \)
\( \frac{2A}{a} = |y_1 - y_2| \)
\( t_1t_2 = -1 \)

Question. PN is an ordinate of the parabola \( y^2 = 4ax \). A straight line is drawn parallel to the axis to bisect NP and meets the curve in Q. NQ meets the tangent at the vertex in a point T such that \( AT = kNP \), then the value of k is (where A is the vertex)
(a) 3/2
(b) 2/3
(c) 1
(d) None of the options
Answer: (b) 2/3
Solution:
\( N(at^2, 0) \)
solve \( y = at \) with curve \( y^2 = 4ax \)
\( x = \frac{at^2}{4} \)
\( Q \left( \frac{at^2}{4}, at \right) \)
Equation of QN: \( y = \frac{at}{\frac{at^2}{4} - at^2} (x - at^2) \)
put \( x = 0 \) \quad \( y = \frac{4}{3} at \)
\( T \left( 0, \frac{4}{3}at \right) \) \quad \( AT = \frac{4}{3}at \)
\( PN = 2at \)
\( \frac{AT}{PN} = \frac{4/3 at}{2at} = \frac{2}{3} \)
so \( k = \frac{2}{3} \)

Question. The tangents to the parabola \( x = y^2 + c \) from origin are perpendicular then c is equal to
(a) 1/2
(b) 1
(c) 2
(d) 1/4
Answer: (d) 1/4
Solution:
\( y^2 = x - c \) ; \( a = 1/4 \)
Slope of tangent = \( \frac{1}{t} \)
so \( \frac{1}{t_1 t_2} = - 1 \)
\( t_1 t_2 = - 1 \dots (i) \)
\( A(a t_1 t_2 + C, a(t_1 + t_2)) \)
\( a t_1 t_2 + C = 0 \)
\( C = - a t_1 t_2 \)
\( C = a \)
\( C = \frac{1}{4} \)
Aliter
\( \frac{c + a + 0}{2} = c \)
\( c + a = 2c \)
\( \implies \) \( c = a \)
\( \implies \) \( c = 1/4 \)

Question. The locus of a point such that two tangents drawn from it to the parabola \( y^2 = 4ax \) are such that the slope of one is double the other is
(a) \( y^2 = \frac{9}{2} ax \)
(b) \( y^2 = \frac{9}{4} ax \)
(c) \( y^2 = 9ax \)
(d) \( x^2 = 4ay \)
Answer: (c) \( y^2 = 9ax \)
Solution:
\( y^2 = 4ax \)
Slope = \( \frac{1}{t} \)
\( \frac{1}{t_1} = \frac{2}{t_2} \)
\( \implies \) \( t_2 = 2t_1 \dots(1) \)
\( R[at_1 t_2, a(t_1 + t_2)] \)
\( h = at_1 t_2 \), \( k = a(t_1 + t_2) \)
\( k = 3at_1 \)
\( \implies \) \( t_1 = \frac{k}{3a} \)
\( h = 2at_1^2 \)
\( h = 2a \frac{k^2}{9a^2} \)
\( \implies \) \( k^2 = \frac{9}{2} ah \)
\( y^2 = \frac{9}{2} ax \)

Question. T is a point on the tangent to a parabola \( y^2 = 4ax \) at its point P. TL and TN are the perpendiculars on the focal radius SP and the directrix of the parabola respectively. Then
(a) SL = 2 (TN)
(b) 3 (SL) = 2 (TN)
(c) SL = TN
(d) 2 (SL) = 3 (TN)
Answer: (c) SL = TN
Solution:
\( \Delta PUT \cong \Delta PLT \)
Both \( \Delta \) are congurrent
Hence \( PU = PL \)
\( PM = SP \)
\( PM - PL = SP - PL \)
\( TN = MU = SL \)

Question. The equation of the circle drawn with the focus of the parabola \( (x - 1)^2 - 8y = 0 \) as its centre and touching the parabola at its vertex is
(a) \( x^2 + y^2 - 4y = 0 \)
(b) \( x^2 + y^2 - 4y + 1 = 0 \)
(c) \( x^2 + y^2 - 2x - 4y = 0 \)
(d) \( x^2 + y^2 - 2x - 4y + 1 = 0 \)
Answer: (d) \( x^2 + y^2 - 2x - 4y + 1 = 0 \)
Solution:
\( (x - 1)^2 = 8y \) ; \( a = 2 \)
\( x - 1 = 0, y = 2 \)
\( x^2 = 8y \) ; \( x = 1, y = 2 \)
vertex \( (1, 0) \)
Focus \( (1, 2) \)
Radius of circle = 2
\( (x - 1)^2 + (y - 2)^2 = 4 \)
\( x^2 + y^2 - 2x - 4y + 1 = 0 \)

Question. Length of the normal chord of the parabola, \( y^2 = 4x \), which makes an angle of \( \frac{\pi}{4} \) with the axis of x is
(a) 8
(b) \( 8\sqrt{2} \)
(c) 4
(d) \( 4\sqrt{2} \)
Answer: (b) \( 8\sqrt{2} \)
Solution:
Normal at \( P(at_1^2, 2at_1) \)
\( a = 1 \)
\( P(t_1^2, 2t_1) \)
\( y + t_1x = 2t_1 + t_1^3 \dots(1) \)
slope = \( 1 = - t_1 \)
\( t_1 = - 1 \)
\( P(1, -2) \) \quad \( t_2 = - t_1 - \frac{2}{t_1} \)
\( Q(t_2^2, 2t_2) \) \quad \( t_2 = 1 + 2 = 3 \)
\( Q(9, 6) \)
\( PQ = \sqrt{(9 - 1)^2 + (6 + 2)^2} = 8\sqrt{2} \)

Question. Tangents are drawn from the point \( (-1, 2) \) on the parabola \( y^2 = 4x \). The length, these tangents will intercept on the line \( x = 2 \)
(a) 6
(b) \( 6\sqrt{2} \)
(c) \( 2\sqrt{6} \)
(d) None of the options
Answer: (b) \( 6\sqrt{2} \)
Solution:
Equation of tangent
\( y = mx + \frac{a}{m} \)
\( a = 1 \)
\( (-1, 2) \)
\( 2 = - m + \frac{1}{m} \)
\( m^2 + 2m - 1 = 0 \)
\( m_1 + m_2 = - 2 \)
\( m_1m_2 = - 1 \)
Length PQ = \( |y_1 - y_2| \)
\( = \left| m_1x + \frac{1}{m_1} - m_2x - \frac{1}{m_2} \right| \) at \( x = 2 \)
\( = \left| 2(m_1 - m_2) - \left( \frac{m_1 - m_2}{m_1m_2} \right) \right| \)
\( = |3(m_1 - m_2)| = 3 \left( 2\sqrt{2} \right) = 6\sqrt{2} \)

Question. Locus of the point of intersection of the perpendicular tangents of the curve \( y^2 + 4y - 6x - 2 = 0 \) is
(a) \( 2x - 1 = 0 \)
(b) \( 2x + 3 = 0 \)
(c) \( 2y + 3 = 0 \)
(d) \( 2x + 5 = 0 \)
Answer: (d) \( 2x + 5 = 0 \)
Solution:
\( y^2 + 4y - 6x - 2 = 0 \)
\( y^2 + 4y + 4 - 6x - 6 = 0 \) ; \( a = \frac{3}{2} \)
\( (y + 2)^2 = 6(x + 1) \)
\( Y^2 = 6X \) vertex \( (-1, -2) \)
POI of tangents \( t_1t_2 = - 1 \)
\( [at_1t_2, a(t_1 + t_2)] \)
\( h + 1 = at_1t_2 \)
\( h + 1 = - \frac{3}{2} \)
\( 2h + 2 = - 3 \)
\( 2h + 5 = 0 \)
\( \implies \) \( 2x + 5 = 0 \)

Question. Tangents are drawn from the points on the line \( x - y + 3 = 0 \) to parabola \( y^2 = 8x \). Then the variable chords of contact pass through a fixed point whose coordinates are
(a) \( (3, 2) \)
(b) \( (2, 4) \)
(c) \( (3, 4) \)
(d) \( (4, 1) \)
Answer: (c) \( (3, 4) \)
Solution:
Let point \( P(x_1, y_1) \)
\( x_1 - y_1 + 3 = 0 \)
C.O.C. w.r.t. \( (x_1, y_1) \) of \( y^2 = 4ax \)
\( yy_1 = 4(x + x_1) \)
\( y(x_1 + 3) = 4x + 4x_1 \)
\( yx_1 + 3y - 4x - 4x_1 = 0 \)
\( (3y - 4x) + x_1(y - 4) = 0 \)
\( L_1 + \lambda L_2 = 0 \)
\( L_1 = 0 \) & \( L_2 = 0 \)
\( 3y = 4x \) \quad \( y = 4 \)
\( x = 3 \)
point \( (3, 4) \)

Question. The line \( 4x - 7y + 10 = 0 \) intersects the parabola, \( y^2 = 4x \) at the points A & B. The co-ordinates of the point of intersection of the tangents drawn at the points A & B are
(a) \( \left( \frac{5}{2}, \frac{7}{2} \right) \)
(b) \( \left( -\frac{7}{2}, \frac{5}{2} \right) \)
(c) \( \left( \frac{7}{2}, \frac{5}{2} \right) \)
(d) \( \left( -\frac{5}{2}, \frac{7}{2} \right) \)
Answer: (a) \( \left( \frac{5}{2}, \frac{7}{2} \right) \)
Solution:
Let : \( 4x - 7y + 10 = 0 \dots(1) \)
C.O.C. w.r.t. \( P(x_1, y_1) \)
\( yy_1 = 2a(x + x_1) \) ; \( a = 1 \)
\( yy_1 = 2(x + x_1) \)
\( 2x - yy_1 + 2x_1 = 0 \dots(2) \)
(1) & (2) are same
\( \frac{4}{2} = \frac{-7}{-y_1} = \frac{10}{2x_1} \)
\( x_1 = \frac{5}{2} \) ; \( y_1 = \frac{7}{2} \) POI \( \left( \frac{5}{2}, \frac{7}{2} \right) \)

Question. From the point \( (4, 6) \) a pair of tangent lines are drawn to the parabola, \( y^2 = 8x \). The area of the triangle formed by these pair of tangent lines & the chord of contact of the point \( (4, 6) \) is
(a) 2
(b) 4
(c) 8
(d) None of the options
Answer: (a) 2
Solution:
\( y^2 = 8x \) ; \( a = 2 \)
Area = \( \frac{(y_1^2 - 8x_1)^{3/2}}{4} \) ; \( (4, 6) \)
\( = \frac{(36 - 32)^{3/2}}{4} = \frac{8}{4} = 2 \) sq. units

Question. TP & TQ are tangents to the parabola, \( y^2 = 4ax \) at P & Q. If the chord PQ passes through the fixed point \( (-a, b) \) then the locus of T is
(a) \( ay = 2b (x - b) \)
(b) \( bx = 2a (y - a) \)
(c) \( by = 2a (x - a) \)
(d) \( ax = 2b (y - b) \)
Answer: (c) \( by = 2a (x - a) \)
Solution:
Equation of PQ
\( (t_1 + t_2)y = 2x + 2at_1t_2 \) passes through \( (-a, b) \)
\( b(t_1 + t_2) = - 2a + 2at_1t_2 \dots(i) \)
\( h = at_1t_2 \) & \( k = a(t_1 + t_2) \)
POI of tangents
\( h = at_1t_2 \) & \( k = a(t_1 + t_2) \)
\( \frac{bk}{a} = - 2a + 2h \)
\( bk = - 2a^2 + 2ah \)
\( by = - 2a^2 + 2ax \)
\( by = 2a(x - a) \)

Question. If the tangent at the point \( P (x_1, y_1) \) to the parabola \( y^2 = 4ax \) meets the parabola \( y^2 = 4a (x + b) \) at Q & R, then the mid point of QR is
(a) \( (x_1 + b, y_1 + b) \)
(b) \( (x_1 - b, y_1 - b) \)
(c) \( (x_1, y_1) \)
(d) \( (x_1 + b, y_1) \)
Answer: (c) \( (x_1, y_1) \)
Solution:
Tangent at P of \( y^2 = 4ax \)
\( yy_1 = 2a(x + x_1) \dots(1) \)
Let Mid point \( (h, k) \)
\( T = S_1 \)
\( yk - 2a(x + h) - 4ab = k^2 - 4a(h + b) \)
\( yk - 2ax - 2ah + 4ah - k^2 = 0 \)
\( yk - 2ax + 2ah - k^2 = 0 \dots(2) \)
(1) & (2) are same
\( \frac{k}{y_1} = \frac{-2a}{-2a} = \frac{2ah - k^2}{-2ax_1} \)
\( k = y_1 \) ; \( -2ax_1 = 2ah - k^2 \)
\( -2ax_1 = 2ah - y_1^2 \) ; \( y_1^2 = 4ax_1 \)
Mid point \( -2ax_1 = 2ah - 4ax_1 \)
\( (x_1, y_1) \) \quad \( 2ah = 2ax_1 \)
\( h = x_1 \)

Question. Let PSQ be the focal chord of the parabola, \( y^2 = 8x \). If the length of SP = 6 then, l(SQ) is equal to (where S is the focus)
(a) 3
(b) 4
(c) 6
(d) None of the options
Answer: (a) 3
Solution:
\( y^2 = 8x \)
\( SP = 6 \)
\( \frac{1}{b} + \frac{1}{c} = \frac{1}{a} \)
\( \frac{1}{c} = \frac{1}{a} - \frac{1}{b} \)
\( c = \frac{ab}{b - a} \) \quad \( b = 6, a = 2 \)
\( = \frac{12}{4} = 3 \)

Question. Two parabolas \( y^2 = 4a(x - l_1) \) and \( x^2 = 4a(y - l_2) \) always touch one another, the quantities \( l_1 \) and \( l_2 \) are both variable. Locus of their point of contact has the equation
(a) \( xy = a^2 \)
(b) \( xy = 2a^2 \)
(c) \( xy = 4a^2 \)
(d) None of the options
Answer: (c) \( xy = 4a^2 \)
Solution:
\( y^2 = 4a(x - l_1) \) \quad \( x^2 = 4a(y - l_2) \)
let the POC \( (h, k) \)
\( 2yy' = 4a \) \quad \( 2x = 4ay' \)
\( y' = \left. \frac{2a}{y} \right|_{(h,k)} = \frac{2a}{k} \dots(1) \)
\( y' = \left. \frac{x}{2a} \right|_{(h,k)} = \frac{h}{2a} \dots(2) \)
(1) and (2) are equal
\( \frac{2a}{k} = \frac{h}{2a} \)
\( hk = 4a^2 \)
\( xy = 4a^2 \)

Question. The straight line joining any point P on the parabola \( y^2 = 4ax \) to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equation of the locus of R is
(a) \( x^2 + 2y^2 - ax = 0 \)
(b) \( 2x^2 + y^2 - 2ax = 0 \)
(c) \( 2x^2 + 2y^2 - ay = 0 \)
(d) \( 2x^2 + y^2 - 2ay = 0 \)
Answer: (b) \( 2x^2 + y^2 - 2ax = 0 \)
Equation of OP:
\( y = \frac{2}{t} x \)
\( k = \frac{2}{t} h \) ....(1)
\( y - 0 = -t(x - a) \)

\( \implies \) \( y = -tx + at \)

\( \implies \) \( k = -th + at \)

\( \implies \) \( \frac{2}{t} h = -th + at \) from (1)
\( \left( t = \frac{2h}{k} \right) \)
\( h = \frac{at^2}{2 + t^2} \)

\( \implies \) \( h = \frac{a \frac{4h^2}{k^2}}{2 + \frac{4h^2}{k^2}} \)

\( \implies \) \( h = \frac{2ah^2}{k^2 + 2h^2} \)

\( \implies \) \( k^2 + 2h^2 = 2ah \)

\( \implies \) \( 2x^2 + y^2 - 2ax = 0 \)

Question. Let A be the vertex and L the length of the latus rectum of parabola, \( y^2 - 2y - 4x - 7 = 0 \). The equation of the parabola with point A as vertex, 2L as the length of the latus rectum and the axis at right angles to that of the given curve is
(a) \( x^2 + 4x + 8y - 4 = 0 \)
(b) \( x^2 + 4x - 8y + 12 = 0 \)
(c) \( x^2 + 4x + 8y + 12 = 0 \)
(d) \( x^2 + 8x - 4y + 8 = 0 \)
Answer: (a) \( x^2 + 4x + 8y - 4 = 0 \), (b) \( x^2 + 4x - 8y + 12 = 0 \)
\( y^2 - 2y - 4x - 7 = 0 \)
\( y^2 - 2y + 1 - 4x - 8 = 0 \)
LR = 4 = L
\( (y - 1)^2 = 4(x + 2) \)
vertex (-2, 1)
Axis = x-axis
New parabola
\( (x + 2)^2 = \pm 8(y - 1) \)
+ve \( (x + 2)^2 = 8(y - 1) \)

\( \implies \) \( x^2 + 4x - 8y + 12 = 0 \)
-ve \( x^2 + 4x + 4 + 8y - 8 = 0 \)

\( \implies \) \( x^2 + 4x + 8y - 4 = 0 \)

Question. Tangent to the parabola \( y^2 = 4ax \) at point P meets the tangents at vertex A at point B and the axis of parabola at T, Q is any point on this tangent and N as the foot of perpendicular from Q on SP, where S is focus, M is the foot of perpendicular from Q on the directrix then
(a) B bisects PT
(b) B trisects PT
(c) QM = SN
(d) QM = 2SN
Answer: (a) B bisects PT, (c) QM = SN
Tangent at P
\( ty = x + at^2 \)
\( B(0, at) \)
\( T(-at^2, 0) \)
clearly B is the mid point of TP

Question. The parametric coordinates of any point on the parabola \( y^2 = 4ax \) can be
(a) \( (at^2, 2at) \)
(b) \( (at^2, -2at) \)
(c) \( (a\sin^2 t, 2a\sin t) \)
(d) \( (a\sin t, 2a\cos t) \)
Answer: (a) \( (at^2, 2at) \), (b) \( (at^2, -2at) \)
\( y^2 = 4ax \)
(a) \( (at^2, 2at) \) possible
(b) \( (at^2, -2at) \) possible
(c) \( (a\sin^2 t, 2a\sin t) \) not possible because \( \sin t \) will lies only in [-1, 1]
so ans. (a), (b)

Question. PQ is a normal chord of the parabola \( y^2 = 4ax \) at P, A being the vertex of the parabola. Through P a line is drawn parallel to AQ meeting the x-axis in R. Then the length of AR is
(a) equal to the length of the latus rectum
(b) equal to the focal distance of the point P.
(c) equal to twice the focal distance of the point P.
(d) equal to the distance of the point P from the directrix
Answer: (c) equal to twice the focal distance of the point P.
Slope of OQ = \( \frac{2}{t_2} \)
line parallel to AQ and passing through P
\( y - 2at_1 = \frac{2}{t_2} (x - at_1^2) \)
For point R put y = 0
\( -2at_1 = \frac{2}{t_2} (x - at_1^2) \)
\( t_2 = -t_1 - \frac{2}{t_1} \)
\( x = at_1^2 - at_1 t_2 \)
\( t_2 + t_1 = -\frac{2}{t_1} \)
\( x = at_1(t_1 - t_2) \)
\( x = 2(at_1^2 + a) = 2 \times \text{focal distance} \)

Question. The length of the chord of the parabola \( y^2 = x \) which is bisected at the point (2, 1) is
(a) \( 5\sqrt{2} \)
(b) \( 4\sqrt{5} \)
(c) \( 4\sqrt{50} \)
(d) \( 2\sqrt{5} \)
Answer: (d) \( 2\sqrt{5} \)
\( T = S_1 \)
\( yy_1 - \frac{1}{2}(x + x_1) = y_1^2 - x_1 \) at \( (x_1, y_1) \implies (2, 1) \)
\( y - \frac{1}{2}(x + 2) = 1 - 2 \)
\( 4y - 2x = 0 \)
\( x = 2y \)

\( \implies \) solve with parabola \( y^2 = x \)
\( y^2 = 2y \)
\( y = 0, y = 2 \)
\( x = 0, x = 4 \)
Points: (0, 0), (4, 2)
\( PQ = \sqrt{4 + 16} = 2\sqrt{5} \)

Question. If the tangents and normals at the extremities of a focal chord of a parabola intersect at \( (x_1, y_1) \) and \( (x_2, y_2) \) respectively, then
(a) \( x_1 = x_2 \)
(b) \( x_1 = y_2 \)
(c) \( y_1 = y_2 \)
(d) \( x_2 = y_1 \)
Answer: (c) \( y_1 = y_2 \)
POI of Two tangents
\( x_1 = at_1 t_2 \)
\( y_1 = a(t_1 + t_2) \)
POI of Two normals
\( x_2 = a(t_1^2 + t_2^2 + t_1 t_2 + 2) \)
\( y_2 = -at_1 t_2(t_1 + t_2) \)
Since it is a focal chord, \( t_1 t_2 = -1 \)
\( y_2 = a(t_1 + t_2) \)
\( y_2 = y_1 \)

Question. Locus of the intersection of the tangents at the ends of the normal chords of the parabola \( y^2 = 4ax \) is
(a) \( (2a + x)y^2 + 4a^3 = 0 \)
(b) \( (x + 2a)y^2 + 4a^2 = 0 \)
(c) \( (x + 2a)y^2 + 4a^3 = 0 \)
(d) None of the options
Answer: (c) \( (x + 2a)y^2 + 4a^3 = 0 \)
\( h = at_1 t_2 \)
\( k = a(t_1 + t_2) \)
For normal chord,
\( t_2 = -t_1 - \frac{2}{t_1} \)

\( \implies \) \( t_2 + t_1 = -\frac{2}{t_1} \)
\( k = a\left(-\frac{2}{t_1}\right) = -\frac{2a}{t_1} \)
\( t_1 = -\frac{2a}{k} \)
\( h = at_1 t_2 = at_1 \left( -t_1 - \frac{2}{t_1} \right) \)

\( \implies \) \( h = - \frac{2a^2}{k} \left( \frac{2a}{k} + \frac{k}{a} \right) \)

\( \implies \) \( h = -\frac{4a^3}{k^2} - 2a \)

\( \implies \) \( hk^2 = -4a^3 - 2ak^2 \)

\( \implies \) \( k^2(h + 2a) + 4a^3 = 0 \)

\( \implies \) \( y^2(x + 2a) + 4a^3 = 0 \)

Question. The locus of the mid point of the focal radii of a variable point moving on the parabola, \( y^2 = 4ax \) is a parabola whose
(a) latus rectum is half the latus rectum of the original parabola
(b) vertex is (a/2, 0)
(c) directrix is y-axis
(d) focus has the co-ordinates (a, 0)
Answer: (a) latus rectum is half the latus rectum of the original parabola, (b) vertex is (a/2, 0), (c) directrix is y-axis, (d) focus has the co-ordinates (a, 0)
Let R(h,k) be the mid point
\( h = \frac{at^2 + a}{2} \)

\( \implies \) \( 2h - a = at^2 \) ....(1)
\( k = \frac{2at + 0}{2} \)

\( \implies \) \( t = \frac{k}{a} \) ....(2)
\( \therefore 2h - a = a\left(\frac{k^2}{a^2}\right) \)

\( \implies \) \( k^2 = 2a\left(h - \frac{a}{2}\right) \)

\( \implies \) \( y^2 = 2a\left(x - \frac{a}{2}\right) \)
vertex = \( \left(\frac{a}{2}, 0\right) \)
directrix : \( x - \frac{a}{2} = -\frac{a}{2} \implies x = 0 \) (y-axis)
Focus : \( x - \frac{a}{2} = \frac{a}{2} \implies x = a \). Focus is (a, 0)

Question. The equation of a straight line passing through the point (3, 6) and cutting the curve \( y = \sqrt{x} \) orthogonally is
(a) 4x + y - 18 = 0
(b) x + y - 9 = 0
(c) 4x - y - 6 = 0
(d) None of the options
Answer: (a) 4x + y - 18 = 0
Equation of Normal In slope form
\( y = mx - 2am - am^3 \); \( a = \frac{1}{4} \)
\( 6 = 3m - \frac{2m}{4} - \frac{m^3}{4} \) at (3, 6)
\( m^3 - 10m + 24 = 0 \)

\( \implies \) \( m = -4 \)
equation of normal
\( y - 6 = -4(x - 3) \)

\( \implies \) \( y + 4x - 18 = 0 \)

Question. The tangent and normal at P (t), for all real positive t, to the parabola \( y^2 = 4ax \) meet the axis of the parabola in T and G respectively, then the angle at which the tangent at P to the parabola is inclined to the tangent at P to the circle through the points P, T and G is
(a) \( \cot^{-1} t \)
(b) \( \cot^{-1} t^2 \)
(c) \( \tan^{-1} t \)
(d) \( \sin^{-1} \left( \frac{t}{\sqrt{1+t^2}} \right) \)
Answer: (c) \( \tan^{-1} t \)
Slope of tangent \( \tan \alpha = 1/t \)
Since \( \angle TPG = 90^\circ \), TG is the diameter of the circle. The tangent to the circle at P is perpendicular to the normal PG. Therefore, the angle between the tangent to the parabola and the tangent to the circle at P is \( \theta \).
\( \tan (90^\circ - \theta) = \cot \theta = \frac{1}{t} \)
\( \tan \theta = t \)
\( \theta = \tan^{-1} t \)

Question. A variable circle is described to passes through the point (1, 0) and tangent to the curve \( y = \tan(\tan^{-1}x) \). The locus of the centre of the circle is a parabola whose
(a) length of the latus rectum is \( 2\sqrt{2} \)
(b) axis of symmetry has the equation x + y = 1
(c) vertex has the co-ordinates (3/4, 1/4)
(d) None of the options
Answer: (b) axis of symmetry has the equation x + y = 1, (c) vertex has the co-ordinates (3/4, 1/4)
\( y = \tan(\tan^{-1}x) = x \)
\( PS = PM \)

\( \implies \) \( (h - 1)^2 + (k - 0)^2 = \frac{(h - k)^2}{2} \)

\( \implies \) \( 2(h^2 + 1 - 2h + k^2) = h^2 + k^2 - 2hk \)

\( \implies \) \( h^2 + k^2 + 2hk + 2 - 4h = 0 \)

\( \implies \) \( x^2 + y^2 + 2xy + 2 - 4x = 0 \)
The focus is S(1,0) and the directrix is \( x - y = 0 \).
Axis of symmetry passes through focus and is perpendicular to directrix: \( x + y = 1 \).
Vertex is the point of intersection of axis and parabola, which is \( (3/4, 1/4) \).

Question. AB, AC are tangents to a parabola \( y^2 = 4ax \). \( p_1 \), \( p_2 \) and \( p_3 \) are the lengths of the perpendiculars from A, B and C respectively on any tangent to the curve, then \( p_2 \), \( p_1 \), \( p_3 \) are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) None of the options
Answer: (b) G.P.
If we consider the tangent at the vertex as our reference tangent:
\( p_2 = at_1^2 \)
\( p_3 = at_2^2 \)
\( p_1 = at_1 t_2 \)
\( p_1 = \sqrt{p_2 p_3} \)
Therefore, \( p_2, p_1, p_3 \) are in G.P.

Question. Through the vertex O of the parabola, \( y^2 = 4ax \) two chords OP and OQ are drawn and the circles on OP and OQ as diameter intersect in R. If \( \theta_1, \theta_2 \) and \( \phi \) are the angles made with the axis by the tangent at P and Q on the parabola and by OR then the value of \( \cot \theta_1 + \cot \theta_2 \) equals
(a) \( -2\tan \phi \)
(b) \( -2\tan(\pi - \phi) \)
(c) 0
(d) \( 2\cot \phi \)
Answer: (a) \( -2\tan \phi \)
\( ty = x + at^2 \)
\( \tan \theta_1 = \frac{1}{t_1}; \tan \theta_2 = \frac{1}{t_2} \)
Circles:
\( (x - at_1^2)(x - 0) + (y - 0)(y - 2at_1) = 0 \)
\( (x - at_2^2)(x - 0) + (y - 0)(y - 2at_2) = 0 \)
For Intersection point R, \( S_1 - S_2 = 0 \)

\( \implies \) \( (at_2^2 - at_1^2)x + y(2at_2 - 2at_1) = 0 \)

\( \implies \) \( 2y + (t_2 + t_1)x = 0 \)

\( \implies \) \( y = - \left( \frac{t_1 + t_2}{2} \right) x \)
\( \tan \theta_1 = \frac{1}{t_1} \)

\( \implies \) \( \cot \theta_1 = t_1 \text{ & } \cot \theta_2 = t_2 \)
\( \cot \theta_1 + \cot \theta_2 = t_1 + t_2 = -2\tan \phi \)

Question. Two parabolas have the same focus. If their directrices are the x-axis & the y-axis respectively, then the slope of their common chord is
(a) 1
(b) -1
(c) 4/3
(d) 3/4
Answer: (b) -1
Let focus be (h, k).
\( C_1 : (x - h)^2 + (y - k)^2 = |y|^2 \)
\( C_2 : (x - h)^2 + (y - k)^2 = |x|^2 \)
\( C_1 - C_2 = 0 \)
\( x^2 - y^2 = 0 \)
\( x^2 = y^2 \)
\( y = \pm x \)