Parabola Notes PDF

Nature of Chapter: 1. You should be comfortable with standard notations of 2 degree curve, before starting this chapter. 2. This chapter has a lot many Results, and all the results are important. 3. Parametric form is very important in Parabola, especially in the formulae of tangents and no...

Nature of Chapter: 1. You should be comfortable with standard notations of 2 degree curve, before starting this chapter. 2. This chapter has a lot many Results, and all the results are important. 3. Parametric form is very important in Parabola, especially in the formulae of tangents and normals. 4. Topic wise practice should be done 2-3 times before going for full chapter solving.  Weightage of Parabola (Last 5 years) 2023 2022 2021 2020 2019 Average JEE Main 1.9 % 2.4 % 2.1 % 2.5 % 2.7 % 2.30 % Jee Advanced 3% 3% 7% 0% 0% 2.60 % Weightage of Conic Sections (Last 5 years) 2023 2022 2021 2020 2019 Average JEE Main 4.3 % 6.8 % 5.0 % 6.8 % 7.5 % 6.08 % Jee Advanced 6% 9% 10 % 6% 3% 6.80 %  Parabola Definition and Standard Equations Parametric form Position of point with respect to parabola Position of line with respect to parabola Focal chord and Focal distance Equations of Normal Chords Some important properties of parabola  Parabola Critical Topics: Basic construction of parabola, ie equations of std parabolas Position of line with respect to parabola Focal chord Results Definition & Standard Equations  Definition A Parabola is the locus of point, which moves in a plane, such that its distance from a fixed point is equal to its distance from a fixed line.  Definition (1) The fixed point S is called focus. directrix (2) The fixed straight line is called the directrix. (3) A line through the focus and perpendicular to the directrix is called the axis (or axis of M P ( h , k) symmetry) of the Parabola. (4) The point of intersection of the Parabola with its axis is called its vertex. S (5) A chord perpendicular to the axis is called focus double ordinate. In particular, the double ordinate through the focus is called the latus rectum. Standard Parabolas & their equations  Standard Parabolas & their equations Here we will be studying parabolas having horizontal or vertical axis.  Standard Parabolas & their equations Standard Parabolas having vertex at the origin Consider the following equations for a > 0 and remember their graphs. (1) y2 = 4ax (2) y2 = −4ax (3) x2 = 4ay (4) x2 = −4ay  Standard Parabolas & their equations Standard Parabolas having vertex at the origin Consider the following equations for a > 0 and remember their graphs. (1) y2 = 4ax (2) y2 = −4ax  Standard Parabolas & their equations Standard Parabolas having vertex at the origin Consider the following equations for a > 0 and remember their graphs. (1) y2 = 4ax (2) y2 = −4ax Y Y LR = 4a LR = 4a Z (−a, 0) O S (a, 0) X S (−a, 0) O X Z (a, 0) x = −a x=a  Standard Parabolas & their equations Standard Parabolas having vertex at the origin (3) x2 = 4ay (4) x2 = −4ay  Standard Parabolas & their equations Standard Parabolas having vertex at the origin (3) x2 = 4ay (4) x2 = −4ay  Find the coordinates of the focus and the vertex, the Q equations of the directrix and the axis, and the length of the latus rectum of the parabola y2 = -12x  Find the coordinates of the focus and the vertex, the Q equations of the directrix and the axis, and the length of the latus rectum of the parabola y2 = -12x Solution:  Find the coordinates of the focus and the vertex, the Q equations of the directrix and the axis, and the length of the latus rectum of the parabola x2 = 6y Solution:  A double ordinate of the parabola y2 = 4ax is of Q length 8a. Prove that the lines from the vertex to its two ends are at right angles.  A double ordinate of the parabola y2 = 4ax is of Q length 8a. Prove that the lines from the vertex to its two ends are at right angles. Solution: Equation of the parabola is : y2 = 4ax... (i) Y A Let AB = 8a be the double ordinate. Then , AM = BM = 4a. M Let OM = x1. Then the coordinates of A and B are X’ X O (x1 , 4a) and (x1 , -4a). Since A lies on (i) B ∴ 16a2 = 4ax1 ⇒ x1 = 4a Y’ ∴ The coordinates of A and B are (4a, 4a) & (4a, -4a). Also, the vertex of (i) is O(0, 0)  A double ordinate of the parabola y2 = 4ax is of Q length 8a. Prove that the lines from the vertex to its two ends are at right angles. Solution: Q The equation of the parabola with vertex at the origin passing through (2, 3) and the axis along x-axis is A 2y2 = 9x B y2 = 9x C y2 = -9x D 2y2 = -9x Q The equation of the parabola with vertex at the origin passing through (2, 3) and the axis along x-axis is A 2y2 = 9x B y2 = 9x C y2 = -9x D 2y2 = -9x Q The equation of the parabola with vertex at the origin passing through (2, 3) and the axis along x-axis is Solution:  Find the equation of parabola with vertex at the origin Q and y + 3 = 0 as its directrix. Also find its focus.  Find the equation of parabola with vertex at the origin Q and y + 3 = 0 as its directrix. Also find its focus. Solution: Let the vertex of the parabola be O(0, 0). Now, y + 3 = 0 ⇒ y = −3. Thus, the directrix is a line parallel to the x-axis at a distance of 3 units below the x-axis. So, the focus of the parabola lies on the y-axis. Consequently, the focus is F(0, 3). Hence, the equation of the parabola is x2 = 4ay, where a = 3 i.e. x2 = 12y  JEE Main 8th Apr, 2023 Q Let A(0, 1), B(1, 1) and C(1, 0) be the midpoints of the sides of a triangle with incentre at D. If the focus of the parabola y2 = 4ax passing through D is , where 𝞪 and 𝞫 are rational numbers, then is equal to A 6 B 8 C 9/2 D 12  JEE Main 8th Apr, 2023 Q Let A(0, 1), B(1, 1) and C(1, 0) be the midpoints of the sides of a triangle with incentre at D. If the focus of the parabola y2 = 4ax passing through D is , where 𝞪 and 𝞫 are rational numbers, then is equal to A 6 B 8 C 9/2 D 12 Solution: D  Standard Parabolas & their equations Standard parabolas having vertex at any point Consider the following equations for a > 0 and remember their graphs. (1) (y − k)2 = 4a(x − h) (2)(y − k)2 = −4a(x − h) (3)(x − h)2 = 4a(y − k) (4)(x − h)2 = −4a(y − k)  Standard Parabolas & their equations Standard parabolas having vertex at any point Consider the following equations for a > 0 and remember their graphs. (1) (y − k)2 = 4a(x − h) (2)(y − k)2 = −4a(x − h)  Standard Parabolas & their equations Standard parabolas having vertex at any point Consider the following equations for a > 0 and remember their graphs. (1) (y − k)2 = 4a(x − h) (2)(y − k)2 = −4a(x − h) LR = 4a LR = 4a A (h, k) A (h, k) Z (h − a, k) S (h + a, k) S (h − a, k) Z (h + a, k) x=h−a x=h+a  Standard Parabolas & their equations Standard parabolas having vertex at any point Consider the following equations for a > 0 and remember their graphs. (3)(x − h)2 = 4a(y − k) (4)(x − h)2 = −4a(y − k) S (h, k + a) Z ( h, k + a ) LR = 4a A ( h, k ) A (h, k) LR = 4a S (h, k − a) Z ( h, k − a )  Draw the following parabola and mark its focus, Q directrix and length of LR : (y + 3)2 = 2(x + 2)  Draw the following parabola and mark its focus, Q directrix and length of LR : (y + 3)2 = 2(x + 2) Solution: Here, a = 1/2 and it is a parabola facing towards right side with vertex (-2, -3) LR = 2 x = - 5/2 y=-3 (-2, -3) S (-3/2, 3)  Draw the following parabola and mark its focus, Q directrix and length of LR : y2 − 2y − 4x + 9 = 0  Draw the following parabola and mark its focus, Q directrix and length of LR : y2 − 2y − 4x + 9 = 0 Solution: The equation y2 − 2y − 4x + 9 = 0 can be written as (y − 1)2 = 4(x − 2) Here, a = 1 and it is a parabola facing towards right side with vertex (2, 1). (2, 1) LR = 4 S (3, 1) x=1  Equation of the parabola whose vertex is (−3, −2), axis is Q horizontal and which passes through the point (1, 2) is A y2 + 4y + 4x − 8 = 0 B y2 + 4y − 4x + 8 = 0 C y2 + 4y − 4x − 8 = 0 D None of these  Equation of the parabola whose vertex is (−3, −2), axis is Q horizontal and which passes through the point (1, 2) is A y2 + 4y + 4x − 8 = 0 B y2 + 4y − 4x + 8 = 0 C y2 + 4y − 4x − 8 = 0 D None of these  Equation of the parabola whose vertex is (−3, −2), axis is Q horizontal and which passes through the point (1, 2) is Solution: Since the axis is horizontal and vertex is (−3, −2), ∴ the equation of the parabola must be of the form (y + 2)2 = 4a(x + 3) It passes through (1, 2), so 16 = 16a i.e., a = 1. Hence, the equation of the required parabola is (y + 2)2 = 4(x + 3) or y2 + 4y − 4x − 8 = 0  Find equations of parabola whose focus and vertex Q are the points (1, 3) and (1, 5) respectively.  Find equations of parabola whose focus and vertex Q are the points (1, 3) and (1, 5) respectively. Solution: Q The vertex of a parabola is (2, 2) and the coordinates of its two extremities of latus rectum are (-2, 0) and (6, 0). Find the equation of the parabola. Q The vertex of a parabola is (2, 2) and the coordinates of its two extremities of latus rectum are (-2, 0) and (6, 0). Find the equation of the parabola. Solution: Shifting the origin at A, equation becomes  If (7, 5) and (7, 3) are the endpoints of the latus Q rectum of a parabola, then find its equation.  If (7, 5) and (7, 3) are the endpoints of the latus Q rectum of a parabola, then find its equation. Solution: Since equation of LR is x = 7, which is a line parallel to the Y-axis, so axis of parabola is parallel to the X-axis (7, 5) (7, 4) Observe that two parabolas are possible here. (7, 3)  If (7, 5) and (7, 3) are the endpoints of the latus Q rectum of a parabola, then find its equation. Solution: (7, 5) (7, 4) (7, 3)  Standard Parabolas & their equations Observation (1) Any horizontal parabola has equation of the form x = ay2 + by + c. (2) Any vertical parabola has equation of the form y = ax2 + bx + c.  Standard Parabolas & their equations Remark The following hold true for all parabolas (i) Distance between vertex and focus  Standard Parabolas & their equations Remark The following hold true for all parabolas (ii) Distance between directrix and Latus rectum  Standard Parabolas & their equations Remark The following hold true for all parabolas (iii) Vertex is always midpoint of focus and point of intersection of axis and directrix.  Standard Parabolas & their equations Remark The following hold true for all parabolas (iv) (Distance of any point on parabola from its axis)2 = Latus Rectum × Distance of that point from tangent at vertex  JEE Main 28th June, 2022 If vertex of a parabola is (2, -1) and the equation of its directrix Q is 4x - 3y = 21, then the length of its latus rectum is A 2 B 8 C 12 D 16  JEE Main 28th June, 2022 If vertex of a parabola is (2, -1) and the equation of its directrix Q is 4x - 3y = 21, then the length of its latus rectum is A 2 B 8 C 12 D 16  JEE Main 28th June, 2022 If vertex of a parabola is (2, -1) and the equation of its directrix Q is 4x - 3y = 21, then the length of its latus rectum is Solution:  Find equation of parabola with vertex (−2, −2) Q and focus (−6, −6).  Find equation of parabola with vertex (−2, −2) Q and focus (−6, −6). Solution: Z (2, 2) A (−2, −2) S (−6, −6)  Find equation of parabola with vertex (−2, −2) Q and focus (−6, −6). Solution: Parametric form of parabola  Parametric form of parabola (1) y2 = 4ax ⇒ P(at2, 2at) (2) x2 = 4ay ⇒ P(2at, at2) (3) (y − k)2 = 4a(x − h) ⇒ P(h + at2, k + 2at) (4)(x − h)2 = 4a(y − k) ⇒ P(h + 2at, k + at2) Remark Try to understand 3 and 4 using shift of origin.  Parametric form of parabola (1a) y2 = 4ax ⇒ P(at2, 2at) (1b) y2 = -4ax ⇒ P(-at2, 2at) (2a) x2 = 4ay ⇒ P(2at, at2) (2b) x2 = -4ay ⇒ P(2at, -at2) Q Let O be the vertex and Q be any point on the parabola x2 = 8y. If the point P divides the line segment OQ internally in the ratio 2 : 3, then the locus of P is ____. Q Let O be the vertex and Q be any point on the parabola x2 = 8y. If the point P divides the line segment OQ internally in the ratio 2 : 3, then the locus of P is ____. Solution: Q Let O be the vertex and Q be any point on the parabola x2 = 8y. If the point P divides the line segment OQ internally in the ratio 2 : 3, then the locus of P is ____. Solution: Q Find the length of the side of an equilateral triangle inscribed in the parabola y2 = 4ax if one of its vertices is at the origin. Q Find the length of the side of an equilateral triangle inscribed in the parabola y2 = 4ax if one of its vertices is at the origin. Solution: Y P (at 2, 2at) y2 = 4ax 30° X O T Q (at 2, −2at) Q Find the length of the side of an equilateral triangle inscribed in the parabola y2 = 4ax if one of its vertices is at the origin. Solution: Q Find the length of the side of an equilateral triangle inscribed in the parabola y2 = 4ax if one of its vertices is at the origin. Alternate Solution: Y P (at 2, 2at) y2 = 4ax 30° X O T Q (at 2, −2at)  Find the locus of middle points of chords of a parabola y2 = 4ax Q which subtend a right angle at the vertex of the parabola. Q Find the locus of middle points of chords of a parabola y2 = 4ax which subtend a right angle at the vertex of the parabola. Solution: Y P (at12, 2at1) (h, k) X O Q (at22, 2at2)  JEE Advanced 2015, P1 Let the curve C be the mirror image of the parabola y2 = 4x Q with respect to the line x + y + 4 = 0. If A and B are the points of intersection of C with the line y = -5, then the distance between A and B is  JEE Advanced 2015, P1 Let the curve C be the mirror image of the parabola y2 = 4x Q with respect to the line x + y + 4 = 0. If A and B are the points of intersection of C with the line y = -5, then the distance between A and B is Ans: 4 Solution: Let P(t2, 2t) be a point on the curve y2 = 4x and Q(h, k) be its image in x + y + 4 = 0. Thus, we have ⇒ h = -(2t + 4) and k = -(t2 + 4) Now, k = -5 So, t = ± 1 Hence, h = -2, -6 So, A and B are (-2, -5) & (-6, -5) Hence, AB = 4  Remark Area of triangle inscribed in a parabola y2 = 4ax Position of a point w.r.t. a Parabola  Position of a point w.r.t. a parabola Result For a parabola whose equation is S = 0 and a given point P(x1, y1), (1) S1 > 0 ⇒ (2) S1 = 0 ⇒ (3) S1 < 0 ⇒  Position of a point w.r.t. a parabola Result For a parabola whose equation is S = 0 and a given point P(x1, y1), (1) S1 > 0 ⇒ point P lies outside the parabola (2) S1 = 0 ⇒ point P lies on the parabola (3) S1 < 0 ⇒ point P lies inside the parabola  The point P(𝞪, 𝞪) lies inside the parabola (y − 2)2 = 4(x − 3) Q for _____. A α ϕ B α (−∞, ∞) C α (−2, 2) D none of these  The point P(𝞪, 𝞪) lies inside the parabola (y − 2)2 = 4(x − 3) Q for _____. A α ϕ B α (−∞, ∞) C α (−2, 2) D none of these  The point P(𝞪, 𝞪) lies inside the parabola (y − 2)2 = 4(x − 3) Q for _____. Solution: Position of a line w.r.t. a Parabola  Position of a line w.r.t. a parabola Recall To ensure that a line L = 0 is tangent to a conic S = 0, we find x or y from L = 0 and put it in S = 0 to get a quadratic and then we take D = 0. Q Find m such that y = mx + 3 is tangent to y2 = 3x. Q Find m such that y = mx + 3 is tangent to y2 = 3x. Solution:  Equation of Tangents to Parabola Various equations of tangents to y2 = 4ax  Equation of Tangents to Parabola Various equations of tangents to y2 = 4ax Slope form Tangent at a point Parametric form on a parabola Y Y slope = m Y (x1, y1) P(t) O X X O O y2 = 4ax y2 = 4ax y2 = 4ax  Equation of Tangents to Parabola Various equations of tangents to y2 = 4ax Slope form Tangent at a point Parametric form on a parabola Y Y slope = m Y (x1, y1) P(t) O X X O O y2 = 4ax y2 = 4ax y2 = 4ax T=0 T=0 ty = x + at2 Q (a) Find m such that y = mx + 3 is tangent to y2 = 3x. (b) Find the equation of the tangents to the parabola y2 = 9x which go through the point (4, 10). Q (a) Find m such that y = mx + 3 is tangent to y2 = 3x. Q (a) Find m such that y = mx + 3 is tangent to y2 = 3x. Solution:  (b) Find the equation of the tangents to the parabola y2 = 9x Q which go through the point (4, 10).  (b) Find the equation of the tangents to the parabola y2 = 9x Q which go through the point (4, 10). Solution:  Equation of Tangents to Parabola Observation From an external point, two tangents can be drawn to a parabola.  IIT 2004 The angle between the tangents drawn from the point Q (1, 4) to the parabola y2 = 4x is ______. A B C D Recall If α and β are roots of ax2 + bx + c = 0, then  IIT 2004 The angle between the tangents drawn from the point Q (1, 4) to the parabola y2 = 4x is ______. A B C D  IIT 2004 The angle between the tangents drawn from the point Q (1, 4) to the parabola y2 = 4x is ______. Solution: Y θ m1 O X m2  If two tangents drawn from the point (𝞪, 𝞫) to the Q hyperbola y2 = 4ax be such that the slope of one tangent is double of the other then A B C D  If two tangents drawn from the point (𝞪, 𝞫) to the Q hyperbola y2 = 4ax be such that the slope of one tangent is double of the other then A B C D Solution:  The equation of the common tangent to the curves Q y2 = 8x and xy = -1 is A 3y = 9x + 2 B y = 2x + 1 C 2y = x + 8 D y=x+2  The equation of the common tangent to the curves Q y2 = 8x and xy = -1 is A 3y = 9x + 2 B y = 2x + 1 C 2y = x + 8 D y=x+2 Solution:  Find the equation of tangent to parabola y = x2 - 2x + 3 at Q point (2, 3).  Find the equation of tangent to parabola y = x2 - 2x + 3 at Q point (2, 3). Solution:  JEE Main 25th July, 2022 Q The sum of diameters of the circles that touch the parabola 75x2 = 64(5y - 3) at the point and the y-axis, is equal to_____.  JEE Main 25th July, 2022 Q The sum of diameters of the circles that touch the parabola 75x2 = 64(5y - 3) at the point and the y-axis, is equal to_____. Ans: 10 Solution: Solution:  NOTE (1) Slope of tangent to y2 = 4ax at P(t) is  NOTE (2) Parametric tangent of x2 = 4ay can be obtained by interchanging x and y in corresponding formula for y2 = 4ax. It does not happen in equation of tangent having given slope m.  NOTE (3) Point of intersection of tangents to y2 = 4ax at P(t1) and Q(t2) is (at1t2, a(t1 + t2)) P(t1) (at1t2, a(t1 + t2)) Q(t2) Q Point of tangency of tangent to y2 = 8x, having slope 2 is __. A B C D Q Point of tangency of tangent to y2 = 8x, having slope 2 is __. A B C D Q Point of tangency of tangent to y2 = 8x, having slope 2 is __. Solution: Q Point of tangency of tangent to y2 = 8x, having slope 2 is __. Alternate Solution:  Tangents to y2 = 4ax make angles θ1 and θ2 with axes. Find locus of Q their point of intersection if cot θ1 + cot θ2 = c, where c is a constant.  Tangents to y2 = 4ax make angles θ1 and θ2 with axes. Find locus of Q their point of intersection if cot θ1 + cot θ2 = c, where c is a constant. Solution: Y θ1 P(t1) θ2 X (h, k) Q(t2) Q Find the equation of the director circle of the parabola y2 = 4ax. Q Find the equation of the director circle of the parabola y2 = 4ax. Solution: Y A (h, k) B(t1) X C(t2) Q Find the equation of the director circle of the parabola y2 = 4ax. Alternate Solution:  NOTE Director circle of a parabola is its directrix. Focal Chords and Focal Distances  Focal Chords and Focal Distances Now, let’s pick focus and study two important terms, namely (1) Focal Distance (2) Focal Chord  Focal Chords and Focal Distances Focal Distance It is the distance of any point on the parabola from its focus. In particular, for y2 = 4ax, Y PS = PM M P ( x1 , y 1 ) O X S x = −a  Focal Chords and Focal Distances Focal Distance It is the distance of any point on the parabola from its focus. In particular, for y2 = 4ax, Y PS = PM M P ( x1 , y 1 ) Eg. If focal distance of a point O X on y2 = 12x is 4 then....... S x = −a  Focal Chords and Focal Distances Focal Distance It is the distance of any point on the parabola from its focus. In particular, for y2 = 4ax, Y PS = PM M P ( x1 , y 1 ) Eg. If focal distance of a point O X on y2 = 12x is 4 then, a + x1 = 4, S that is 3 + x1 = 4 ⇒ x1 = 1 So, the point is x = −a  Focal Chords and Focal Distances Focal Chord Any chord passing through the focus of a parabola is called a focal chord. We should be remembering some results related to focal chords Result (1) For y2 = 4ax, if P(t1) and Q(t2) are the endpoints of a focal chord then t1t2 = − 1. (2) Tangents at endpoints of a focal chord are perpendicular and hence intersect on directrix. (3) Circle with endpoints of focal chord as diameter, always touches directrix  Result (4) Length of a focal chord of y2 = 4ax, making an angle α with the X-axis, is 4acosec2α.  Result (5) Length of focal chord whose one endpoint is P(t) is a(t + 1/t)2 (6) If AB is a focal chord of y2 = 4ax, then , where S is the focus.  JEE Main 9th Jan, 2020 Q If one end of focal chord AB of the parabola y2 = 8x is at then the equation of the tangent to it at B is: A x - 2y + 8 = 0 B x + 2y + 8 = 0 C 2x - y -24 = 0 D 2x + y -24 = 0  JEE Main 9th Jan, 2020 Q If one end of focal chord AB of the parabola y2 = 8x is at then the equation of the tangent to it at B is: A x - 2y + 8 = 0 B x + 2y + 8 = 0 C 2x - y -24 = 0 D 2x + y -24 = 0  JEE Main 9th Jan, 2020 Q If one end of focal chord AB of the parabola y2 = 8x is at then the equation of the tangent to it at B is: Solution:  JEE Main 2019 Q If one end of a focal chord of the parabola, y2 = 16x is at (1, 4), then the length of this focal chord is A 24 B 20 C 22 D 25  JEE Main 2019 Q If one end of a focal chord of the parabola, y2 = 16x is at (1, 4), then the length of this focal chord is A 24 B 20 C 22 D 25  JEE Main 2019 Q If one end of a focal chord of the parabola, y2 = 16x is at (1, 4), then the length of this focal chord is Solution:  JEE Advanced 2013 Q Let PQ be a focal chord of the parabola y2 = 4ax. The tangents to the parabola at P and Q meet at a point lying on the line y = 2x + a, a > 0. Length of chord PQ is : A 7a B 5a C 2a D 3a  JEE Advanced 2013 Q Let PQ be a focal chord of the parabola y2 = 4ax. The tangents to the parabola at P and Q meet at a point lying on the line y = 2x + a, a > 0. Length of chord PQ is : A 7a B 5a C 2a D 3a Solution:  JEE Main 1st Feb, 2023 If the x-intercept of a focal chord of the parabola Q y2 = 8x + 4y + 4 is 3, then the length of this chord is equal to____.  JEE Main 1st Feb, 2023 If the x-intercept of a focal chord of the parabola Q y2 = 8x + 4y + 4 is 3, then the length of this chord is equal to____. Answer: 16  JEE Main 1st Feb, 2023 If the x-intercept of a focal chord of the parabola Q y2 = 8x + 4y + 4 is 3, then the length of this chord is equal to____. Solution: Normals of a parabola  Normals of a parabola First we try to observe that the equation of normal can be easily developed from parametric tangent. Y P(t) X  Normals of a parabola First we try to observe that the equation of normal can be easily developed from parametric tangent. Y Normal at P(t) is given by y − 2at = − t(x + at2), P(t) i.e. y + tx = 2at + at3 (parametric normal of y2 = 4ax) X As t = − m, so we also have y = mx − 2am − am3 (normal of y2 = 4ax having slope m) Q Find the equation of the normal to y2 = 4x, having slope of 3. Also find its point of contact. Q Find the equation of the normal to y2 = 4x, having slope of 3. Also find its point of contact. Solution:  JEE Main 9th Apr, 2017 Q If y = mx + c is the normal at a point on the parabola y2 = 8x whose focal distance is 8 units, then |c| is equal to: A B C D  JEE Main 9th Apr, 2017 Q If y = mx + c is the normal at a point on the parabola y2 = 8x whose focal distance is 8 units, then |c| is equal to: A B C D Solution:  Number of distinct normal that can be drawn to the parabola Q y2 = 4x from the point is: A 1 B 2 C 3 D 4  Number of distinct normal that can be drawn to the parabola Q y2 = 4x from the point is: A 1 B 2 C 3 D 4  Number of distinct normal that can be drawn to the parabola Q y2 = 4x from the point is: Solution:  Normals of a parabola NOTE For x2 = 4ay, interchange x and y to get its parametric normal, that is x + ty = 2at + at3 i.e. so, we have is normal to x2 = 4ay having slope m.  The equation of the normal to the parabola x2 = 8y Q at x = 4 is: A x + 2y = 0 B x+y=2 C x - 2y =0 D x+y=6  The equation of the normal to the parabola x2 = 8y Q at x = 4 is: A x + 2y = 0 B x+y=2 C x - 2y =0 D x+y=6 Solution: Given equation of parabola : x2 = 8y ∴a = 2 As we know that, parametric equation of normal to the parabola x2 = 4ay at t is x + ty = 2at + at3 ∴ Equation of normal to the given parabola is : x + ty = 4t + 2t3 Point of normality is (2at, t2) As we have to find normal at x = 4 ∴ 2at = 4 ⇒2x2xt=4 ⇒t=1 ∴ Equation of normal is: x + y = 4 + 2 i.e. x + y = 6  Normals of a parabola NOTE For y2 = 4ax, the normal having slope m is Y y = mx − 2am − am3. m1 If it passes through the point (h, k), then P ( h, k ) X Q R m2 m3  Normals of a parabola NOTE For y2 = 4ax, the normal having slope m is Y y = mx − 2am − am3. m1 If it passes through the point (h, k), then P am3 +(2a − h)m + k = 0. So, at most 3 normals can pass through a ( h, k ) given point, and slopes of these normals X m1, m2, m3 satisfy : Q R m1 + m2 + m3 = 0, m2 m3  Normals of a parabola NOTE We can also develop relations between parameters of P, Q, R using the fact that mN = − t. So, we have : t1 + t2 + t3 = 0,  Normals of a parabola Remark The points P, Q, R, normals at which are concurrent, are called conormal points.  Observations Centroid of triangle formed by 3 conormal points of y2 = 4ax lies on the X-axis. Q Let the normals at points A(4a, -4a) and B(9a, -6a) on the parabola y2 = 4ax meet the point P. The equation of the normal from P on y2 = 4ax (other than PA and PB) is A 5x + y - 135a = 0 B 5x - y + 115a = 0 C 5x + y + 115 = 0 D 5x - y - 115a = 0 Q Let the normals at points A(4a, -4a) and B(9a, -6a) on the parabola y2 = 4ax meet the point P. The equation of the normal from P on y2 = 4ax (other than PA and PB) is A 5x + y - 135a = 0 B 5x - y + 115a = 0 C 5x + y + 115 = 0 D 5x - y - 115a = 0 Q Let the normals at points A(4a, -4a) and B(9a, -6a) on the parabola y2 = 4ax meet the point P. The equation of the normal from P on y2 = 4ax (other than PA and PB) is Solution: Let, A = (at12, 2at1), B = (at22, 2at2) And the foot of the required normal is (at32, 2at3), then 2at1 = -4a, 2at2 = -6a and t1 + t2 + t3 = 0 t1 = -2, t2 = -3 and -2 - 3 + t3 = 0 t3 = 5 Hence, the equation of the required normal is y = -t3x + 2at3 + at32 y = -5x + 10a + 125a y + 5x - 135a = 0  Find the locus of the point of intersection of two Q perpendicular normals to y2 = 4ax. Solution: Q The locus of a point such that two of the three normals drawn from it to the parabola y2 = 4ax coincide is A 27ay2 = 4(x + 2a)3 B 27ay2 = -4(x - 2a)3 C 27ay2 = 4(x - 2a)3 D 27ay2 = -4(x + 2a)3 Q The locus of a point such that two of the three normals drawn from it to the parabola y2 = 4ax coincide is A 27ay2 = 4(x + 2a)3 B 27ay2 = -4(x - 2a)3 C 27ay2 = 4(x - 2a)3 D 27ay2 = -4(x + 2a)3 Solution: Let y = mx - 2am - am3 be the equation of normal Since it passes through (h, k) ∴ am3 + (2a - h)m + k = 0 ⇒ 𝚺m1 = 0 …(i) ⇒ …(ii) ⇒ …(iii) Since m2 = m3 ∴ from (i) we have m1 = -2m2 …(iv) From (iii) we have m1m22 = -k/a …(v) From (iv) and (v), …(vi) Also am23 + (2a - h)m2 = -k …(vii) From (vi) and (vii) we have 4(h - 2a)3 = 27ak2  Normals of a parabola Result (1) If normal to y2 = 4ax at P(t1) intersects the parabola again at Q(t2), then  Normals of a parabola Result (2) If normals to y2 = 4ax at P(t1) and Q(t2) intersect on the parabola at R(t3), then (a) t1t2 = 2 (b) t3 = −(t1 + t2)  Normals of a parabola Result (3) 3 distinct normals drawn to y2 = 4ax from (h, 0) are real if h > 2a. Q The normal to parabola y2 = 4x at P(4, 4), intersects parabola again at Q. Find equation of tangent at Q. Solution: Slide 177 1 add ex: Normal to ysq = 4x at P(4,4), intersects parabola again at Q. Find equation of tangent at Q. Arvind Kalia, 13-09-2023  JEE Main 16th Mar, 2021 Q If the three normals drawn to the parabola, y2 = 2x pass through the point (a, 0), a ≠ 0, then ‘a’ must be greater than: A B C -1 D 1  JEE Main 16th Mar, 2021 Q If the three normals drawn to the parabola, y2 = 2x pass through the point (a, 0), a ≠ 0, then ‘a’ must be greater than: A B C -1 D 1  JEE Main 16th Mar, 2021 Q If the three normals drawn to the parabola, y2 = 2x pass through the point (a, 0), a ≠ 0, then ‘a’ must be greater than: Solution: Chords of a parabola  Chords of a parabola The formulae for the equation of the chord of contact and a chord with given midpoint remain the same, that is T = 0 and T = S1 respectively.  (a) 3x + y = 6 intersects y2 = 4x at A and B. Find the Q point of intersection of the tangents at A and B. (b) Find the locus of the midpoints of the chords of y2 = 4x that pass through the focus.  (a) 3x + y = 6 intersects y2 = 4x at A and B. Find the Q point of intersection of the tangents at A and B. Solution:  (b) Find the locus of the midpoints of the chords of Q y2 = 4x that pass through the focus.  (b) Find the locus of the midpoints of the chords of Q y2 = 4x that pass through the focus. Solution: Q If the chord of contact of tangents from a point P to the parabola y2 = 4ax touches the parabola x2 = 4by, then find the locus of P. Solution: Chord of contact of parabola y2 = 4ax w.r.t. Point P(x1, y1) is yy1 = 2a(x + x1)... (i) This line touches the parabola x2 = 4by Solving Eq. (i) with parabola, we have or y1x2 - 8abx - 8abx1 = 0 According to the question, this equation must have equal roots. ⇒ D=0 ⇒ 64a2b2 + 32abx1y1 = 0 ⇒ x1y1 = -2ab or xy = -2ab Which is a rectangular hyperbola. Some important Properties of Parabola  Some important properties of parabola Here, we will be studying some properties which hold for all parabolas. We can prove these properties for y2 = 4x easily.  Result (1) The portion of a tangent to a parabola, intercepted between the directrix and the point of tangency, subtends a right angle at the focus. (2) Foot of perpendicular from the focus of a parabola, on the tangent at any point on the parabola, lies on tangent at vertex.  Result (3) The tangent at any point P on a parabola bisects the angle between the focal chord through P and the perpendicular from P to the directrix.  Result (4) Reflection property : A ray of light coming parallel to the axis of a parabola, passes through the focus after getting reflected from inside the parabola.  Result 5(a) Area of triangle made by the points P(t1), Q(t2), R(t3) on the parabola with equation y2 = 4ax is given by the magnitude of the determinant , that is |a2 (t1 - t2)(t2 - t3)(t3 - t1)|.  Result 5(b) Area of triangle made by 3 points on a parabola is twice the area of triangle made by the tangents at these 3 points.  JEE Advanced 2021, P2 Let E denote the parabola y2 = 8x. Let P = (-2, 4) and let Q and Q Q’ be two distinct points on E such that the lines PQ and PQ’ are tangents to E. Let F be the focus of E. Then which of the following statements is (are) TRUE ? The triangle PFQ is a right-angled A triangle The triangle QPQ’ is a right-angled B triangle C The distance between P and F is 5√2 D F lies on the line joining Q and Q’  JEE Advanced 2021, P2 Let E denote the parabola y2 = 8x. Let P = (-2, 4) and let Q and Q Q’ be two distinct points on E such that the lines PQ and PQ’ are tangents to E. Let F be the focus of E. Then which of the following statements is (are) TRUE ? The triangle PFQ is a right-angled A triangle The triangle QPQ’ is a right-angled B triangle C The distance between P and F is 5√2 D F lies on the line joining Q and Q’ Solution:  IIT JEE 2011 Q Consider the parabola y2 = 8x. Let Δ1 be the area of the triangle formed by the end points of its latus rectum and the point on the parabola, and Δ2 be the area of the triangle formed by drawing tangents at P and at the end points of the latus rectum. Then is  IIT JEE 2011 Q Consider the parabola y2 = 8x. Let Δ1 be the area of the triangle formed by the end points of its latus rectum and the point on the parabola, and Δ2 be the area of the triangle formed by drawing tangents at P and at the end points of the latus rectum. Then is Ans: 2 Solution: