Parabola JEE Mathematics Worksheets Set 01

Question. In the parabola \( y^2 = 4ax \), the tangent at the point P, whose abscissa is equal to the latus ractum meets the axis in T & the normal at P cuts the parabola again in Q. Prove that \( PT : PQ = 4 : 5 \).
Answer: Point P is \( (4a, 4a) \)
equation of tangent at P :
\( x - 2y + 4a = 0 \) .....(i)
& equation of normal at P
\( 2x + y = 12a \) .....(ii)
point T = \( (-4a, 0) \)
Q = \( (9a, -6a) \)
\( PT : PQ = 4 : 5 \)

Question. Two tangents to the parabola \( y^2 = 8x \) meet the tangent at its vertex in the points P & Q. If PQ = 4 units, prove that the locus of the point of the intersection of the two tangents is \( y^2 = 8 (x + 2) \).
Answer: Equation of pair of tangent through the point (h, k)
\( SS_1 = T^2 \)
\( \Rightarrow (y^2 - 8x) (k^2 - 8h) = (ky - 4(x + h))^2 \) ......(i)
solving with y-axis i.e. x = 0
\( - 8hy^2 + 8khy - 16h^2 = 0 \)
Now \( |y_1 - y_2| = 4 = d \)
locus
\( \Rightarrow y^2 = 8(x + 2) \)

Question. Two straight lines one being a tangent to \( y^2 = 4ax \) and the other to \( x^2 = 4by \) are at right angles. Find the locus of their point of intersection.
Answer: Let the equation of tangent to \( y^2 = 4ax \) is
\( y = mx + a/m \) .......(i)
then let the equation that is tangent to \( x^2 = 4by \) has slope \( -1/m \)
so equation of tangent to \( x^2 = 4by \)
\( y = -\frac{1}{m}x - \frac{1}{m^2}b \) .......(ii)
let the point of intersection of (i) & (ii) is (h, k)
so locus ; \( (ax + by) (x^2 + y^2) + (bx - ay)^2 = 0 \).

Question. A variable chord PQ of the parabola \( y^2 = 4x \) is drawn parallel to the line \( y = x \). If the parameters of the points P & Q on the parabola are p & q respectively, show that \( p + q = 2 \). Also show that the locus of the point of intersection of the normals at P & Q is \( 2x - y = 12 \).
Answer: equation of chord through \( t_1 \) & \( t_2 \)
\( 2x - (t_1 + t_2)y + 2at_1 t_2 = 0 \)
Now slope = 1
\( \Rightarrow \frac{2}{t_1 + t_2} = 1 \)
\( \Rightarrow t_1 + t_2 = 2 \) (H.P.) ...(i)
point of intersection of normals at \( t_1 \) & \( t_2 \)
\( (h, k) = (a(t_1^2 + t_2^2 + t_1 t_2 + 2), -at_1 t_2 (t_1 + t_2)) \)
so locus is \( 2x - y = 12 \)

Question. Show that an infinite number of triangles can be inscribed in either of the parabolas \( y^2 = 4ax \) & \( x^2 = 4by \) whose sides touch the other.
Answer: Let the sides of \( \Delta \) touches \( y^2 = 4ax \), so their
equation are \( y = m_1 x + \frac{a}{m_1} \), \( y = m_2 x + \frac{a}{m_2} \) & \( y = m_3 x + \frac{a}{m_3} \)
vertices of \( \Delta = \left(\frac{a}{m_1 m_2}, a\left(\frac{1}{m_1} + \frac{1}{m_2}\right)\right) \), \( \left(\frac{a}{m_2 m_3}, a\left(\frac{1}{m_2} + \frac{1}{m_3}\right)\right) \) & \( \left(\frac{a}{m_3 m_1}, a\left(\frac{1}{m_1} + \frac{1}{m_3}\right)\right) \)
Let two of these lie on \( x^2 = 4by \), then we have to prove that remaining lies on the curve.
for two vertices ; \( \frac{a^2}{(m_1 m_2)^2} = 4ab \left(\frac{1}{m_1} + \frac{1}{m_2}\right) \) ...(i)
& \( \frac{a^2}{(m_2 m_3)^2} = 4ab \left(\frac{1}{m_2} + \frac{1}{m_3}\right) \) .....(ii)
subtracting (i) & (ii) & using (i)
\( \frac{a^2}{(m_3 m_1)^2} = 4ab \left(\frac{1}{m_3} + \frac{1}{m_1}\right) \)
(similar to (i) & (ii))
so 3rd vertex also lie on the \( x^2 = 4by \)

Question. If \( (x_1, y_1) \), \( (x_2, y_2) \) and \( (x_3, y_3) \) be three points on the parabola \( y^2 = 4ax \) and the normals at these points meet in a point then prove that
\( \frac{x_1 - x_2}{y_3} + \frac{x_2 - x_3}{y_1} + \frac{x_3 - x_1}{y_2} = 0 \).
Answer: \( 2ay + y_1 x = 2ay_1 + x_1 y_1 \) .......(i)
at \( (x_2, y_2) \) : \( 2ay + y_2 x = 2ay_2 + x_2 y_2 \) .......(ii)
at \( (x_3, y_3) \) : \( 2ay + y_3 x = 2ay_3 + x_3 y_3 \) .......(iii)
(i), (ii) & (iii) are concurrent so
\[ \begin{vmatrix} 2a & y_1 & 2ay_1 + x_1 y_1 \\ 2a & y_2 & 2ay_2 + x_2 y_2 \\ 2a & y_3 & 2ay_3 + x_3 y_3 \end{vmatrix} = 0 \]

Question. Prove that the parabola \( y^2 = 16x \) and the circle \( x^2 + y^2 - 40x - 16y - 48 = 0 \) meet at the point P(36, 24) and one other point Q. Prove that PQ is a diameter of the circle. Find Q.
Answer: Center of circle = (20, 8)
which is outside the parabola
Point Q is (4, -8) of diameter
(If (36, 24) is one point & (20, 8) is center then using it other point of diameter is (4, -8) & so on)

Question. A variable tangent to the parabola \( y^2 = 4ax \) meets the circle \( x^2 + y^2 = r^2 \) at P and Q. Prove that the locus of the mid point of PQ is \( x(x^2 + y^2) + ay^2 = 0 \).
Answer: Equation of tangent to parabola \( y^2 = 4ax \)
\( y = mx + a/m \) ......(i)
solving with circle \( x^2 + y^2 = r^2 \)
\( x^2 (1 + m^2) + 2ax + a^2/m^2 - r^2 = 0 \) ......(ii)
Let the mid point of P & Q is (h, k)
\( h = \frac{x_1 + x_2}{2} = \frac{-a}{1 + m^2} \) ......(iii)
{\( \because x_1 \) & \( x_2 \) are roots of (ii)}
Similarly \( k = \frac{a}{m(1 + m^2)} \) ......(iv)
using (iii) & (iv) locus of (h, k) is \( x(x^2 + y^2) + ay^2 = 0 \)

Question. Show that the locus of the centroids of equilateral triangles inscribed in the parabola \( y^2 = 4ax \) is the parabola \( 9y^2 - 4ax + 32a^2 = 0 \).
Answer: Let vertices of the triangle be \( t_1, t_2 \) & \( t_3 \)
so centroid is (h, k)
Now \( h = \frac{a(t_1^2 + t_2^2 + t_3^2)}{3} \) .....(i)
& \( k = \frac{2a(t_1 + t_2 + t_3)}{3} \) .....(ii)
also if slope of one side is \( \frac{2}{t_1 + t_2} = m_1 \) (let) &
\( \frac{2}{t_2 + t_3} = m_2 \) (let)
then \( \tan 60^\circ = \frac{m_2 - m_1}{1 + m_1 m_2} \) .....(iii)
using (i), (ii) & (iii), locus of centroid is
\( 9y^2 - 4ax + 32a^2 = 0 \)

Question. Show that the normals at the points \( (4a, 4a) \) & at the upper end of the latus rectum of the parabola \( y^2 = 4ax \) intersect on the same parabola.
Answer: \( y^2 = 4ax \)
\( P(4a, 4a) \quad L.R. \text{ extreme} \)
\( L_1 (a, 2a) \)
Let \( P(at_1^2, 2at_1) \quad L_1(at_2^2, 2at_2) \)
\( 2at_1 = 4a \)
\( \implies t_1 = 2 \)
\( 2at_2 = 2a \)
\( \implies t_2 = 1 \)
POI of two normals
\( [a (t_1^2 + t_2^2 + t_1 t_2 + 2), -at_1 t_2(t_1 + t_2)] \)
\( [9a, -6a] \)
which satisfy \( y^2 = 4ax \) parabola

Question. Find the equations of the tangents to the parabola \( y^2 = 16x \), which are parallel & perpendicular respectively to the line \( 2x - y + 5 = 0 \). Find also the coordinates of their points of contact.
Answer: \( y^2 = 16x \)
\( \implies 4a = 16 \)
\( \implies a = 4 \)
Let the POC \( (at^2, 2at) \)
Tangent
\( ty = x + at^2 \)
slope \( = \frac{1}{t} = 2 \) and \( \frac{1}{t} = \frac{-1}{2} \)
\( t = \frac{1}{2} \quad t = -2 \)
POC \( (1, 4) \) \quad POC \( (16, -16) \)
equation
\( \frac{y}{2} = x + \frac{a}{4} \quad \text{equation} \)
\( y = 2x + 2 \quad -2y = x + 4 \times 4 \)
\( 2y + x + 16 = 0 \)

Question. Find the equations of the tangents of the parabola \( y^2 = 12x \), which passes through the point \( (2, 5) \).
Answer: \( y^2 = 12x \)
\( \implies 4a = 12 \)
\( \implies a = 3 \)
POI of tangent \( [3t_1 t_2, 3(t_1 + t_2)] \)
\( 3t_1 t_2 = 2 \) ......(1)
\( 3(t_1+ t_2) = 5 \) ......(2)
\( t_1 - t_2 = \sqrt{(t_1 + t_2)^2 - 4t_1 t_2} = \sqrt{\frac{25}{9} - \frac{8}{3}} \)
\( t_1 - t_2 = \frac{1}{3} \) ......(3)
By (2) and (3)
\( t_1 = 1, \quad t_2 = \frac{2}{3} \)
Tangent
\( ty = x + at^2 \) and
\( ty = x + 3t^2 \quad \frac{2}{3}y = x + 3 \times \frac{4}{9} \)
\( y = x + 3 \quad 2y = 3x + 4 \)

Question. Through the vertex O of a parabola \( y^2 = 4x \), chords OP & OQ are drawn at right angles to one another. Show that for all positions of P, PQ cuts the axis of the parabola at a fixed point. Also find the locus of the middle point of PQ.
Answer: \( P(at_1^2, 2at_1) \quad a = 1 \)
\( P(t_1^2, 2t_1) \)
\( Q(t_2^2, 2t_2) \)
\( m_{AP} = \frac{2}{t_1} \quad m_{AQ} = \frac{2}{t_2} \)
\( \frac{2}{t_1} \times \frac{2}{t_2} = - 4 \)
\( \implies t_1 t_2 = - 4 \)
Fixed point \( (4, 0) \)
Let the middle point \( M(h, k) \)
\( h = \frac{t_1^2 + t_2^2}{2} \)
\( \implies t_1^2 + t_2^2 = 2h \) ......(1)
\( k = \frac{2t_1 + 2t_2}{2} \)
\( \implies t_1 + t_2 = k \) ......(2)
From (1) \( (t_1 + t_2)^2 - 2t_1 t_2 = 2h \)
\( k^2 + 8 = 2h \)
\( k^2 = 2(h - 4) \)
\( \implies y^2 = 2(x - 4) \)

Question. Three normals to \( y^2 = 4x \) pass through the point \( (15, 12) \). Show that if one of the normals is given by \( y = x - 3 \) & find the equations of the others.
Answer: \( y^2 = 4x \)
normal equation in slope form
\( y = mx - 2am - am^3 \) ; \( a = 1 \)
\( y = mx - 2m - m^3 \)
passes through \( (15, 12) \)
\( 12 = 15m - 2m - m^3 \)
\( m^3 - 13m + 12 = 0 \)
\( \implies m = 1, m = - 4, m = 3 \)
normals are
\( m = 1 \)
\( \implies y = x - 3 \)
\( m = - 4 \)
\( \implies y = - 4x + 72 \)
\( m = 3 \)
\( \implies y = 3x - 33 \)

Question. If the normal at \( P(18, 12) \) to the parabola \( y^2 = 8x \) cuts it again at Q, show that \( 9PQ = 80\sqrt{10} \)
Answer: \( y^2 = 8x \) ; \( a = 2 \)
\( P(18, 12) \)
\( P(at_1^2, 2at_1) \)
\( P(2t_1^2, 4t_1) \)
\( 2t_1^2 = 18 \)
\( t_1 = \pm 3 \)
\( t_1 = 3 \quad t_1 = - 3 \)
\( P(18, 12) \quad t_2 = - t_1 - \frac{2}{t_1} \)
\( t_2 = -3 - \frac{2}{3} = -\frac{11}{3} \)
\( Q \left( \frac{242}{9}, -\frac{44}{3} \right) \quad P(18, 12) \)
\( PQ = \frac{80}{9} \sqrt{10} \)
\( \implies 9PQ = 80\sqrt{10} \)

Question. Prove that, the normal to \( y^2 = 12x \) at \( (3,6) \) meets the parabola again in \( (27, -18) \) & circle on this normal chord as diameter is \( x^2 + y^2 - 30x + 12y - 27 = 0 \).
Answer: \( y^2 = 12x \quad a = 3 \)
\( (3, 6) \)
\( 2at_1 = 6 \)
\( t_1 = 1 \)
\( t_2 = - t_1 - \frac{2}{t_1} = - 3 \)
\( (at_2^2, 2at_2) = (27, - 18) \)
equation of circle
\( (x - 3) (x - 27) + (y - 6) (y + 18) = 0 \)
\( x^2 + y^2 - 30x + 12y - 27 = 0 \)

Question. P & Q are the points of contact of the tangents drawn from the point T to the parabola \( y^2 = 4ax \). If PQ be the normal to the parabola at P, prove that TP is bisected by the directrix.
Answer: \( t_2 = - t_1 - \frac{2}{t_1} \)
\( \implies t_2 + t_1 = - \frac{2}{t_1} \) ....(1)
M is the mid point of PT
\( M(h, k) \)
\( h = \frac{at_1^2 + at_1 t_2}{2} = \frac{at_1(t_1 + t_2)}{2} = \frac{at_1}{2} \left( -\frac{2}{t_1} \right) \)
\( T(at_1 t_2, a(t_1 + t_2)) \)
\( h = - a \)
\( x = - a \)

Question. From the point \( (-1, 2) \) tangent lines are drawn to the parabola \( y^2 = 4x \). Find the equation of the chord of contact. Also find the area of the triangle formed by the chord of contact & the tangents.
Answer: \( y^2 = 4x \)
C.O.C. \( T = 0 \)
\( yy_1 = 2a(x + x_1) \)
\( 2y = 2(x - 1) \)
\( y = x - 1 \)
area \( = \frac{(y_1^2 - 4x_1)^{3/2}}{2a} = \frac{(4 + 4)^{3/2}}{2} = 8\sqrt{2} \)

Read the information given and answer the question 18, 19, 20.
From the point \( P(h, k) \) three normals are drawn to the parabola \( x^2 = 8y \) and \( m_1 \), \( m_2 \) and \( m_3 \) are the slopes of three normals.

Question. Find the algebraic sum of the slopes of these three normals.
Answer: \( x^2 = 4by \)
Normal equation
\( y = mx + 2b + \frac{b}{m^2} \)
Here \( b = 2 \)
\( (h, k) \)
\( k = mh + 4 + \frac{2}{m^2} \) ......(1)
\( hm^3 + (4 - k)m^2 + 2 = 0 \)
\( m_1 + m_2 + m_3 = \frac{k - 4}{h} \)
\( m_1 m_2 m_3 = -\frac{2}{h} \)

Question. If two of the three normals are at right angles then the locus of point P is a conic, find the latus rectum of conic.
Answer: \( m_1 m_2 m_3 = -\frac{2}{h} \)
\( m_1 m_2 = - 1 \)
\( m_3 = \frac{2}{h} \)
put \( m_3 \) in equation (1)
\( k = 2 + 4 + \frac{2}{4 /h^2} \)
\( k = 6 + \frac{h^2}{2} \)
\( h^2 = 2(k - 6) \)
\( x^2 = 2(y - 6) \)
\( LR = 4a = 2 \)

Question. If the two normals from P are such that they make complementary angles with the axis then the locus of point P is a conic, find a directrix of conic.
Answer: \( m_1 m_2 = 1 \)
\( m_3 = -\frac{2}{h} \)
put in equation (1)
\( k = - 2 + 4 + \frac{2}{4 /h^2} \)
\( k = 2 + \frac{h^2}{2} \)
\( h^2 = 2(k - 2) \)
\( x^2 = 2(y - 2) \)
Directrix : \( Y = - a \)
\( y - 2 = -\frac{1}{2} \)
\( y = 2 - \frac{1}{2} = \frac{3}{2} \)
\( 2y - 3 = 0 \)