7 Most FAQ’s of Parabola Chapter in Inter 2nd Year Maths-2B (TS/AP)
7 Marks
LAQ-1 : Derive the standard form of the parabola
Let S be the focus and L = 0 be the directrix of the parabola.
Let Z be the projection of S onto the directrix.
Let A be the midpoint of SZ implying $$SA = AZ$$
$$\Rightarrow SA/AZ = 1$$ indicating A is a point on the parabola.
Taking AS as the principal axis of the parabola as the X-axis and the line perpendicular to AS through A as the Y-axis,
$$\Rightarrow A = (0,0)$$
Let $$AS = a$$
$$\Rightarrow S = (a,0) Z = (-a,0)$$
The equation of the directrix is $$x = -a$$
$$\Rightarrow x + a = 0$$
Let $$P(x_1, y_1)$$ be any point on the parabola,
N be the projection of P onto the Y-axis,
M be the projection of P onto the directrix.
Here $$PM = PN + NM = x_1 + a$$
Now, by the focus-directrix property of the parabola, we have $$SP/PM = 1$$
$$\Rightarrow SP = PM$$
$$\Rightarrow SP^2 = PM^2$$
$$\Rightarrow (x_1 – a)^2 + (y_1 – 0)^2 = (x_1 + a)^2$$
$$\Rightarrow y_1^2 = (x_1 + a)^2 – (x_1 – a)^2$$
$$\Rightarrow y_1^2 = 4ax_1$$
The equation of the locus of $$P(x_1,y_1)$$ is $$y^2 = 4ax$$
LAQ-2 : Find the equation of the parabola passing through the points (-1,2),(1,-1),(2,1) and having its axis parallel to the x-axis
Given the equation of the parabola whose axis is parallel to the x-axis is $$x = ly^2 + my + n$$
The parabola passes through $$(-1,2)$$
$$\Rightarrow -1 = l(2^2) + m(2) + n \Rightarrow 4l + 2m + n = -1 \quad …(1)$$
The parabola passes through $$(1,-1)$$
$$\Rightarrow 1 = l(-1)^2 + m(-1) + n \Rightarrow l – m + n = 1 \quad …(2)$$
The parabola passes through $$(2,1)$$
$$\Rightarrow 2 = l(1)^2 + m(1) + n \Rightarrow l + m + n = 2 \quad …(3)$$
From (1) – (2) we get
$$3l + 3m = -2$$
$$\Rightarrow l + m = -2/3 \quad …(4)$$
From (2) – (3) we get
$$-2m = -1$$
$$\Rightarrow m = 1/2$$
From (4) substituting $$m = 1/2$$ gives
$$l + 1/2 = -2/3$$
$$\Rightarrow l = -7/6$$
Substituting $$l = -7/6$$ and $$m = 1/2$$ back into the equations to find n, we reassess the calculation to ensure accuracy.
Using (3) to find n, substituting l and m,
$$l + m + n = 2 \Rightarrow -7/6 + 1/2 + n = 2$$
$$\Rightarrow n = 2 + 7/6 – 1/2 = 2 + 7/6 – 3/6 = 2 + 4/6 = 2 + 2/3 = 8/3$$
Hence, the correct value of n is 8/3, and the equation of the required parabola is:
$$x = -\frac{7}{6}y^2 + \frac{1}{2}y + \frac{8}{3}$$
Thus, the equation simplifies to: $$7y^2 – 3y – 16 – 6x = 0$$
LAQ-3 : Find the equation of the parabola whose axis is parallel to the x-axis and passing through the points (-2,1), (1,2) and (-1,3)
The equation of the parabola whose axis is parallel to the x-axis is $$x = ly^2 + my + n$$
The parabola passes through $$(-2,1)$$
$$-2 = l(1^2) + m(1) + n \Rightarrow l + m + n = -2 \quad…(1)$$
The parabola passes through $$(1,2)$$
The parabola passes through $$(-1,3)$$
$$-1 = l(3^2) + m(3) + n \Rightarrow 9l + 3m + n = -1 \quad…(3)$$
Subtracting (1) from (2) gives: $$3l + m = 3 \quad…(4)$$
Subtracting (2) from (3) gives: $$5l + m = -2 \quad…(5)$$
Solving (4) and (5) for l and m:
$$2l = -5 \Rightarrow l = -\frac{5}{2}$$
Substituting $$l = -\frac{5}{2}$$ in (4) gives:
$$m = \frac{21}{2}$$
Using $$l = -\frac{5}{2}$$ and $$m = \frac{21}{2}$$ in (1) to find n:
$$-\frac{5}{2} + \frac{21}{2} + n = -2 \Rightarrow n = -10$$
Substituting the values $$l = -\frac{5}{2} m = \frac{21}{2} n = -10$$ in $$x = ly^2 + my + n$$ we get the equation of the required parabola as:
$$x = -\frac{5}{2}y^2 + \frac{21}{2}y – 10$$
Simplifying: $$5y^2 – 21y + 2x + 20 = 0$$
LAQ-4 : Find the equation of the parabola whose axis is parallel to the y-axis and passing through the points (4,5),(-2,11),(-4,21)
Given the equation of the parabola whose axis is parallel to the y-axis is $$y = lx^2 + mx + n$$
The parabola passes through $$(4,5)$$
$$5 = 16l + 4m + n \quad …(1)$$
The parabola passes through $$(-2,11)$$
$$11 = 4l – 2m + n \quad …(2)$$
The parabola passes through $$(-4,21)$$
$$21 = 16l – 4m + n \quad …(3)$$
Subtracting (2) from (1) gives: $$12l + 6m = -6 \quad …(4)$$
Subtracting (2) from (3) gives: $$12l – 2m = 10 \quad …(5)$$
Solving (4) and (5) for m: $$8m = -16 \Rightarrow m = -2$$
Now substituting $$m = -2$$ into (4) gives: $$12l + 6(-2) = -6 \Rightarrow 12l = 6 \Rightarrow l = \frac{1}{2}$$
Substituting $$l = \frac{1}{2} m = -2$$ into (1) to find n :
$$16\left(\frac{1}{2}\right) + 4(-2) + n = 5 \Rightarrow 8 – 8 + n = 5 \Rightarrow n = 5$$
Substituting the values $$l = \frac{1}{2} m = -2 n = 5$$ in $$y = lx^2 + mx + n$$ we have:
$$y = \frac{1}{2}x^2 – 2x + 5$$
Simplified, the equation becomes: $$x^2 – 4x + 10 – 2y = 0$$
LAQ-5 : Show that the equations of common tangents to the circle x2+y2=2a2 and the parabola y2=8ax are y = ±(x+2a)
Given parabola is $$y^2 = 8ax$$
$$\Rightarrow y^2 = 4(2a)x \quad …(1)$$
The equation of the tangent to (1) with slope m is $$y = mx + \frac{2a}{m} \quad …(2)$$
Comparing the above equation with $$y = mx + c$$ we get $$c = \frac{2a}{m}$$
Given circle is $$x^2 + y^2 = 2a^2 \quad …(3)$$
Applying the tangential condition $$c^2 = r^2(1+m^2)$$ between (2) and (3) we get
$$\left(\frac{2a}{m}\right)^2 = 2a^2(1+m^2)$$
$$\Rightarrow \frac{4a^2}{m^2} = 2a^2(1+m^2)$$
$$\Rightarrow \frac{2}{m^2} = (1+m^2)$$
$$\Rightarrow (1+m^2)m^2 = 2$$
$$\Rightarrow m^4 + m^2 = 2$$
$$\Rightarrow m^2 = 1$$
$$\Rightarrow m = ±1$$
From (2), the equations of the required common tangents are $$y = ±x + \frac{2a}{±1}$$
$$y = x + 2a$$
LAQ-6 : If y1,y2,y3 are the y-coordinates of the vertices of the triangle inscribed in the parabola y2=4ax then show that the area of the triangle is 1/8a|(y1-y2)(y2-y3)(y3-y1)| sq. units
We take vertices as $$P(x_1,y_1) = (at_1^2,2at_1) Q(x_2,y_2) = (at_2^2,2at_2) R(x_3,y_3) = (at_3^2,2at_3)$$
The area of $$\Delta PQR = \frac{1}{2}\left| \begin{array}{cc} x_1 – x_2 & x_1 – x_3 \ y_1 – y_2 & y_1 – y_3 \end{array} \right|$$
$$= \frac{1}{2}\left| \begin{array}{cc} at_1^2 – at_2^2 & at_1^2 – at_3^2 \ 2at_1 – 2at_2 & 2at_1 – 2at_3 \end{array} \right|$$
$$= \frac{a \cdot 2a}{2}\left| \begin{array}{cc} t_1^2 – t_2^2 & t_1^2 – t_3^2 \ t_1 – t_2 & t_1 – t_3 \end{array} \right|$$
$$= a^2\left| \begin{array}{cc} (t_1 – t_2)(t_1 + t_2) & (t_1 – t_3)(t_1 + t_3) \ t_1 – t_2 & t_1 – t_3 \end{array} \right|$$
$$= a^2(t_1 – t_2)(t_1 – t_3)\left| \begin{array}{cc} t_1 + t_2 & t_1 + t_3 \ 1 & 1 \end{array} \right|$$
$$= a^2(t_1 – t_2)(t_1 – t_3)\left| t_1 + t_2 – t_1 – t_3 \right|$$
$$= a^2 \left| (t_1 – t_2)(t_2 – t_3)(t_3 – t_1) \right|$$
$$= a^2\left| \left(\frac{y_1}{2a} – \frac{y_2}{2a}\right)\left(\frac{y_2}{2a} – \frac{y_3}{2a}\right)\left(\frac{y_3}{2a} – \frac{y_1}{2a}\right) \right|$$
$$= \frac{a^2}{(2a)^3}\left| (y_1 – y_2)(y_2 – y_3)(y_3 – y_1) \right| = \frac{1}{8a}\left| (y_1 – y_2)(y_2 – y_3)(y_3 – y_1) \right|$$
LAQ-7 : Prove that the area of the triangle formed by the tangents at (x1,y1),(x2,y2),(x3,y3) to the parabola y2=4ax is 1/16a|(y1-y2)(y2-y3)(y3-y1)| sq. units
Let $$D(x_1,y_1) = (at_1^2, 2at_1) E(x_2,y_2) = (at_2^2, 2at_2) F(x_3,y_3) = (at_3^2, 2at_3)$$
The equations of the tangents at D E and F on the parabola $$y^2 = 4ax$$
$$t_1y = x + at_1^2$$
$$t_2y = x + at_2^2$$
$$t_3y = x + at_3^2$$
Solving the above equations in pairs to get the points of intersection:
From (1) and (2), we solve for y:
$$(t_1 – t_2)y = a(t_1^2 – t_2^2)$$
$$\Rightarrow y = a(t_1 + t_2)$$
Using this y value in (1), solve for x:
$$x = t_1y – at_1^2 = at_1(t_1 + t_2) – at_1^2 = at_1t_2$$
Point of intersection of tangents at D and E is $$P = (at_1t_2, a(t_1 + t_2))$$
Similarly, for E and F and F and D we get:
$$Q = (at_2t_3, a(t_2 + t_3))$$
$$R = (at_3t_1, a(t_3 + t_1))$$
The area of $$\Delta PQR$$ is given by:
$$= \frac{1}{2}\begin{vmatrix} at_1t_2 – at_2t_3 & at_1t_2 – at_3t_1 \ a(t_1 + t_2) – a(t_2 + t_3) & a(t_1 + t_2) – a(t_3 + t_1) \end{vmatrix}$$
$$= \frac{a^2}{2}\begin{vmatrix} t_2(t_1 – t_3) & t_1(t_2 – t_3) \ t_1 – t_3 & t_2 – t_3 \end{vmatrix}$$
$$= \frac{a^2}{2}(t_1 – t_3)\begin{vmatrix} t_2 & t_1 \ 1 & 1 \end{vmatrix}$$
$$= \frac{a^2}{2}(t_1 – t_3)(t_2 – t_3)(t_2 – t_1)$$
$$= \frac{a^2}{2}\left|\left(\frac{y_1}{2a} – \frac{y_2}{2a}\right)\left(\frac{y_2}{2a} – \frac{y_3}{2a}\right)\left(\frac{y_3}{2a} – \frac{y_1}{2a}\right)\right|$$